MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 2
Liberty University
Question 1
Question
Suppose that a certain investment doubles in value every 6 years.
If you initially invest $5000, how much will your investment be worth after
18 years?
Solution
Step 1: Determine the growth factor. Since the investment doubles in value
every 6 years, the growth factor is 2.
Step 2: Calculate the number of compounding periods. Since the investment
grows every 6 years, and we are interested in the value after 18 years, there will
be 18/6 = 3 compounding periods.
Step 3: Use the compound interest formula A=P(1 + r
n)nt
, where: - A
is the final amount - Pis the principal amount (initial investment) - ris the
interest rate (growth factor - 1) - nis the number of compounding periods per
year - tis the total number of years
Step 4: Plug in the values into the formula:
A= 5000 ×(1 + 1
1)1×3
Step 5: Simplify the expression and calculate the final amount:
A= 5000 ×23= 5000 ×8 = $40000
Therefore, after 18 years, the investment will be worth $40000.
Question 2
Question
Solve the exponential equation 3x+1 −2·3x= 5.
Solution
Step 1: Let’s simplify the given exponential equation by factoring out 3x:
3x+1 −2·3x= 3x·31−2·3x
= 3 ·3x−2·3x
= (3 −2) ·3x
= 1 ·3x
= 3x
Step 2: After simplifying, our equation becomes 3x= 5.
Step 3: To solve 3x= 5, we can take the natural logarithm of both sides to
eliminate the exponential:
ln(3x) = ln(5)
Step 4: By the logarithmic property ln(ab)=bln(a), we have:
xln(3) = ln(5)
Step 5: Solve for xby dividing both sides by ln(3):
x=ln(5)
ln(3)
Therefore, the solution to the exponential equation 3x+1 −2·3x= 5 is
x=ln(5)
ln(3) .
Question 3
Question
Let f(x) = 3 ·2x+2. Find the equation of the asymptote for y= log2(f(x)).
Solution
Step 1: Rewrite f(x)using the properties of exponents.
f(x) = 3 ·2x+2
= 3 ·2x·22
= 12 ·2x
2
Step 2: Now, find log2(f(x)).
log2(f(x)) = log2(12 ·2x)
= log2(12) + log2(2x)
= log2(12) + x
Step 3: The equation of the asymptote for y= log2(f(x)) is y=x.
Question 4
Question
Suppose a certain investment grows according to the formula A(t) = P ert , where
A(t)represents the amount of money in the account after tyears, Pis the initial
investment, ris the annual interest rate (expressed as a decimal), and tis the
time in years.
A student invests $5000 in an account that offers 6How long will it take for
the investment to double in value?
Solution
Let’s first identify the given information: - Initial investment, P= $5000 -
Annual interest rate, r= 0.06 - Compounded continuously implies the formula
A(t) = P ert
We are looking for the time tit will take for the investment to double in
value. This means we are looking for the time tat which A(t) = 2P.
Step 1: Substitute the given values into the formula A(t) = P ert and set it
equal to 2P.
2P=P ert
Step 2: Cancel out the Pterms on both sides of the equation.
2 = ert
Step 3: Take the natural logarithm of both sides to solve for t.
ln 2 = ln(ert )
ln 2 = rt
Step 4: Divide both sides by rto solve for t.
t=ln 2
r
t=ln 2
0.06
3
Step 5: Calculate the value of t.
t≈0.693147
0.06
t≈11.55245
Therefore, it will take approximately 11.55 years for the investment to double
in value.
Question 5
Question
Samantha invested $10,000 in a savings account that yields an annual interest
rate of 4.5%, compounded continuously. How much money will Samantha have
in her account after 10 years?
Solution
Step 1: The formula for continuous compound interest is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount ($10,000 in this case), - ris the annual interest rate (4.5% or 0.045),
and - tis the time the money is invested for (10 years).
Step 2: Substitute the given values into the formula and solve for A:
A= 10000 ·e0.045·10
Step 3: Simplify the exponential expression:
A= 10000 ·e0.45
Step 4: Evaluate e0.45 using a calculator:
A≈10000 ·1.57029
Step 5: Calculate the final amount in Samantha’s account after 10 years:
A≈$15,702.90
Therefore, Samantha will have approximately $15,702.90 in her account after
10 years.
Question 6
Question
A population of bacteria in a petri dish is modeled by the function P(t) =
100e0.05t, where tis the time in hours. Find the rate of growth of the population
after 3 hours.
4
Solution
Step 1: To find the rate of growth of the population after 3 hours, we need to
find the derivative of the population function P(t)with respect to time t.
Step 2: Let P′(t)denote the rate of growth of the population at time t. We
have:
P′(t) = d
dt(100e0.05t)
Step 3: Using the chain rule of differentiation, we have:
P′(t) = 100 ·0.05e0.05t
Step 4: Simplifying the expression, we get:
P′(t) = 5e0.05t
Step 5: To find the rate of growth after 3 hours, we evaluate P′(3):
P′(3) = 5e0.05(3)
Step 6: Calculate the value:
P′(3) = 5e0.15 ≈5(1.161) ≈5.805
Step 7: Therefore, the rate of growth of the population after 3 hours is
approximately 5.805 bacteria per hour.
Question 7
Question
Samantha invests $10,000 in an account that earns 5% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Step 1: Express the continuous compound interest formula.
The continuous compound interest formula is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears - Pis the principal
amount (initial investment) - ris the annual interest rate (in decimal form) - e
is the base of the natural logarithm (approximately equal to 2.71828) - tis the
time the money is invested or borrowed for
Step 2: Set up the equation for doubling the initial investment.
When the initial investment doubles, the amount becomes 2P. Based on the
continuous compound interest formula, this occurs when:
2P=P·e0.05t
5
Step 3: Solve for the time tit takes for the investment to double.
Divide both sides of the equation by Pto isolate the exponential term:
2 = e0.05t
Step 4: Solve for tusing logarithms.
To solve for t, take the natural logarithm of both sides:
ln(2) = ln(e0.05t)
Step 5: Use the properties of logarithms to simplify.
The natural logarithm of the base eis just 1, so the equation simplifies to:
ln(2) = 0.05t
Step 6: Solve for t.
Divide by 0.05 to isolate t:
t=ln(2)
0.05
Step 7: Calculate the exact time it will take for the investment to double.
Substitute the values into the formula:
t=ln(2)
0.05 ≈0.6931
0.05 ≈13.86 years
Therefore, it will take approximately 13.86 years for Samantha’s investment
to double in value.
Question 8
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 can be written as 33.
Step 2: Set up the equation with the same base.
3x−1= 33.
Step 3: Equate the exponents.
x−1 = 3.
Step 4: Solve for x.
x= 3 + 1
6
x= 4.
Step 5: Check the solution by substituting x= 4 back into the original
equation.
34−1= 27
33= 27
27 = 27.
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 9
Question
Suppose a certain investment grows according to the exponential function A(t) =
5000 ·1.07t, where A(t)represents the amount of money in the account after t
years. Determine the time it will take for the initial investment of 5000totripleinvalue.
Solution
Step 1: Set up the equation for when the initial investment triples in value.
5000 ·1.07t= 3 ·5000
Step 2: Simplify the equation.
1.07t= 3
Step 3: Take the natural logarithm of both sides to solve for t. Recall that
if ex=y, then ln(y) = x. Applying this to our equation:
ln(1.07t)= ln(3)
tln(1.07) = ln(3)
Step 4: Solve for t.
t=ln(3)
ln(1.07)
Step 5: Using a calculator, calculate the value of t.
t≈ln(3)
ln(1.07) ≈1.09861
0.06303 ≈17.44
Step 6: Therefore, it will take approximately 17.44 years for the initial in-
vestment of 5000totripleinvalue.
7
Question 10
Question
Solve the exponential equation: 52x−1= 125.
Solution
Step 1: Rewrite both sides of the equation using the same base.
52x−1= 125
52x−1= 53
Step 2: Since the bases are the same, we can set the exponents equal to each
other.
2x−1 = 3
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 4: Check the solution.
52(2)−1= 125
53= 125
125 = 125
Therefore, the solution to the exponential equation 52x−1= 125 is x= 2.
Question 11
Question
Suppose a certain investment grows according to the exponential function A(t) =
5000 ×1.06twhere A(t)represents the amount of money invested after tyears.
(a) Find the initial investment.
(b) After how many years will the investment double?
(c) Determine the amount of money after 10 years if the investment is left
untouched.
8
Solution
(a) To find the initial investment, we look for the value of A(0). Substitute
t= 0 into the exponential function:
A(0) = 5000 ×1.060
A(0) = 5000 ×1
A(0) = 5000
Therefore, the initial investment was $5000.
(b) To find when the investment will double, we need to solve for tin the
equation A(t) = 2 ×A(0).
5000 ×1.06t= 2 ×5000
1.06t= 2
To solve for t, take the natural logarithm of both sides:
ln(1.06t)= ln(2)
tln(1.06) = ln(2)
t=ln(2)
ln(1.06) ≈11.89
Therefore, the investment will double after approximately 11.89 years.
(c) To determine the amount of money after 10 years, we can simply substi-
tute t= 10 into the exponential function:
A(10) = 5000 ×1.0610 ≈5000 ×1.790847 ≈8954.24
After 10 years, the amount of money in the investment will be approximately
$8954.24.
Question 12
Question
Samantha invests $5000 in a savings account that earns 3% annual interest
compounded quarterly. How much will Samantha have in her account after 10
years?
9
Solution
Step 1: First, we find the quarterly interest rate by dividing the annual interest
rate by 4. In this case, the quarterly interest rate is r=3%
4= 0.75% = 0.0075.
Step 2: We use the compound interest formula to determine the amount in
the account after 10 years:
A=P(1 + r
n)nt
where: - Ais the amount in the account after tyears, - Pis the principal amount
(initial investment), - ris the quarterly interest rate, - nis the number of times
interest is compounded per year, and - tis the number of years.
Step 3: Substituting the given values into the formula, we have:
A= 5000 (1 + 0.0075
4)4×10
Step 4: Simplifying, we get:
A= 5000 ×(1.001875)40
Step 5: Calculating the value inside the parentheses gives:
A= 5000 ×1.354035786
Step 6: Finally, we find:
A≈6770.18
Therefore, Samantha will have approximately $6770.18 in her account after
10 years.
Question 13
Question
Suppose a population of bacteria doubles every 3 hours. If there are initially
100 bacteria in the population, find an exponential model for the population
size after thours.
Solution
Step 1: Let’s denote the initial population of bacteria as P0= 100 and the
growth rate as r(since the population doubles every 3 hours, r= 2).
Step 2: The exponential growth formula is given by P(t) = P0·(1 + r)t
3,
where P(t)represents the population size at time t.
Step 3: Substitute P0= 100,r= 2, and simplify the equation to find the
exponential model for the population size after thours:
P(t) = 100 ·(1 + 2) t
3= 100 ·3t
3= 100 ·3t
3
Therefore, the exponential model for the population size after thours is
P(t) = 100 ·3t
3.
10
Question 14
Question
Let e2x·ex= 48. Find the exact value of x.
Solution
Step 1: Simplify the left side of the equation. We know that ea·eb=ea+b.
Therefore,
e2x·ex=e2x+x=e3x.
Step 2: Replace e3xin the original equation.
e3x= 48.
Step 3: Use the definition of the natural logarithm (ln) to solve for x. Take
the natural logarithm of both sides of the equation.
ln(e3x)= ln(48).
Step 4: Recall that ln(ea) = afor any real number a. Therefore, we have:
3x= ln(48).
Step 5: Finally, solve for xby dividing both sides by 3.
x=ln(48)
3.
So, the exact value of xis ln(48)
3.
Question 15
Question
The population of a town is modeled by the function P(t) = 2000 ·1.03twhere
trepresents time in years. Find the population after 10 years and the annual
growth rate of the population.
Solution
Step 1: To find the population after 10 years, plug in t= 10 into the population
function.
P(10) = 2000 ·1.0310
P(10) = 2000 ·1.344031
P(10) = 2688.062
So, the population after 10 years is 2688.062.
Step 2: The annual growth rate, which is the factor by which the population
increases each year, is given by the multiplier in the exponential function. In
this case, the growth rate is 1.03 or 3
Therefore, the population grows at an annual rate of 3
11
Question 16
Question
Suppose the population of a town is modeled by the function P(t) = 5000·1.03t,
where trepresents time in years.
How long will it take for the population of the town to double?
Solution
Step 1: Set up the equation for population doubling.
We need to find the time tit takes for the population to double, which means
P(t) = 2 ·P(0), where P(0) is the initial population.
2·5000 = 5000 ·1.03t
Step 2: Solve for t.
Divide both sides by 5000:
2 = 1.03t
Step 3: Take the natural logarithm of both sides.
ln(2) = ln(1.03t)
Step 4: Use the property of logarithms ln(ab)=b·ln(a)to simplify the right
side.
ln(2) = t·ln(1.03)
Step 5: Solve for tby dividing both sides by ln(1.03).
t=ln(2)
ln(1.03)
Step 6: Calculate the approximate value of t.
t≈0.6931
0.02956 ≈23.44
Therefore, it will take approximately 23.44 years for the population of the
town to double.
Question 17
Question
Solve the exponential equation 23x= 16.
12
Solution
Step 1: Rewrite 16 as a power of 2.
16 = 24
Step 2: Substitute 16 as 24in the equation 23x= 16.
23x= 24
Step 3: Equate the exponents.
3x= 4
Step 4: Solve for xby dividing both sides by 3.
x=4
3
Therefore, the solution to the equation 23x= 16 is x=4
3.
Question 18
Question
Solve the exponential equation 42x−1= 64.
Solution
Step 1: Rewrite 4and 64 in terms of the same base. Step 2: Solve the resulting
exponential equation for x.
Step 1: First, rewrite both 4and 64 with the base 4:
42x−1= 64
(22)2x−1= 43
22(2x−1) = 26
24x−2= 26
Step 2: Since the bases are the same, we can equate the exponents:
4x−2 = 6
4x= 8
x= 2
Therefore, the solution to the equation 42x−1= 64 is x= 2.
13
Question 19
Question
Solve the equation 32x= 125 for x.
Solution
Step 1: Rewrite 125 as a power of 3. Since 125 = 33, the equation becomes
32x= 33.
Step 2: Set the exponents equal to each other: 2x= 3.
Step 3: Solve for xby dividing both sides by 2: x=3
2.
Therefore, the solution to the equation 32x= 125 is x=3
2.
Question 20
Question
Sara invests $5,000 in an account that earns 8% interest compounded annually.
How much money will be in the account after 10 years? Round your answer to
the nearest dollar.
Solution
Step 1: First, we will use the formula for compound interest to find the future
value of the investment:
A=P(1 + r
n)nt
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment) of $5,000, - ris the annual interest rate of 8% or
0.08, - nis the number of times the interest is compounded per year (in this
case, once annually), - tis the number of years the money is invested for.
Step 2: Substitute the given values into the formula:
A= 5000 (1 + 0.08
1)1∗10
Step 3: Calculate the future value of the investment:
A= 5000 ×(1 + 0.08)10
A= 5000 ×(1.08)10
Step 4: Calculate (1.08)10:
(1.08)10 ≈2.158924
14
Step 5: Substitute the approximation back into the equation:
A= 5000 ×2.158924
A≈10794.62
Step 6: Round the answer to the nearest dollar: After 10 years, the amount
of money in the account will be approximately $10,795.
Question 21
Question
Solve the following exponential equation for x:32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the two exponential expressions equal to each other.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
Step 4: Solve for x.
2x= 2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation:
32(1)+1 = 27
33= 27
27 = 27
Step 6: Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 22
Question
A certain investment grows exponentially. If the initial investment was $10,000
and it doubled in value after 10 years, determine the growth rate of the invest-
ment.
15
Solution
Let Pbe the initial investment, rbe the growth rate, and tbe the time in years.
Step 1: Use the formula for exponential growth to set up the equation.
Given that the investment doubles in value after 10 years, we have:
2P=P×ert
Step 2: Plug in the given values. We know that the initial investment
P= $10,000 and the time t= 10 years. Substituting these values into the
equation, we get:
2(10000) = 10000 ×e10r
Step 3: Solve for the growth rate r. Dividing both sides by 10000, we get:
2 = e10r
Taking the natural logarithm of both sides to solve for r, we get:
ln(2) = ln(e10r)
ln(2) = 10r
r=ln(2)
10
Step 4: Calculate the growth rate r. Using a calculator, we find
r≈ln(2)
10 ≈0.0693
Therefore, the growth rate of the investment is approximately 0.0693, or
6.93
Question 23
Question
Solve the exponential equation: 3x+1 −3x= 20.
Solution
Step 1: Use the properties of exponents to simplify the equation. Step 2: Let
u= 3x. Step 3: Substitute uinto the equation. Step 4: Solve for u. Step 5:
Substitute back to solve for x.
Step 1: Simplify the equation using the properties of exponents:
3x+1 −3x= 20
3·3x−3x= 20
16
3·3x−3x= 20
Step 2: Let u= 3x. Step 3: Substitute uinto the equation:
3u−u= 20
2u= 20
Step 4: Solve for u:
u=20
2
u= 10
Step 5: Substitute back to solve for x:
3x= 10
x= log3(10)
Question 24
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
We know that 27 = 33, so the equation becomes 3x−1= 33.
Step 2: Set the exponents equal to each other.
Since the bases are the same, we can set the exponents equal: x−1 = 3.
Step 3: Solve for x.
Add 1to both sides to solve for x:x= 4.
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 25
Question
Solve the exponential equation 32x−3= 27.
Solution
Step 1: Rewrite 27 using base 3. Step 2: Solve for x.
17
Question 26
Question
Solve the exponential equation 2·32x−1= 6x+ 18 for x.
Solution
Step 1: Rewrite both sides with the same base. Step 2: Use the properties of
exponents to simplify the equation. Step 3: Solve the resulting equation for x.
Step 1: Rewrite both sides with the same base.
We can rewrite 6as 32, so the given equation becomes:
2·32x−1= (32)x+ 18
Step 2: Use the properties of exponents to simplify the equation.
Now, simplify the right side:
2·32x−1= 32x+ 18
Step 3: Solve the resulting equation for x.
So, the equation we need to solve is:
2·32x−1= 32x+ 18
Let’s now substitute 32xas afor simplification:
2·32x−1=a+ 18
2·32x−1=a+ 18
2·32x−1=a+ 18
So, the equation to solve is:
2·32x−1=a+ 18
Can you figure out how to solve it now?
Question 27
Question
The population of a town is modeled by the function P(t) = 5000×1.03t, where
trepresents the number of years since the year 2020.
27. Determine the population of the town in the year 2030 according to the
model.
18
Solution
Step 1: First, we need to determine the value of tfor the year 2030. Since t
represents the number of years since 2020, in the year 2030, t= 2030−2020 = 10.
Step 2: Substitute t= 10 into the function P(t)to find the population in
the year 2030.
P(10) = 5000 ×1.0310
Step 3: Calculate 1.0310:
1.0310 ≈1.344031
Step 4: Multiply 5000 by 1.344031 to find the population in the year 2030:
P(10) ≈5000 ×1.344031 ≈6720.155
Therefore, according to the model, the population of the town in the year
2030 is approximately 6720.
Question 28
Question
Solve the following exponential equation:
5x−1= 125
Solution
Step 1: Rewrite 125 as a power of 5:
125 = 53
Step 2: Substitute the expression for 125 into the equation:
5x−1= 53
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
x−1 = 3
Step 4: Solve for x:
x= 3 + 1
x= 4
Step 5: Therefore, the solution to the exponential equation is x= 4.
19
Question 29
Question
Solve the following exponential equation for x:32x= 81.
Solution
Step 1: Rewrite 81 as a power of 3.
81 = 34
Step 2: Set the two expressions equal to each other.
32x= 34
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x= 4
Step 4: Solve for x.
x=4
2
x= 2
Step 5: Therefore, the solution to the equation 32x= 81 is x= 2.
Question 30
Question
Solve the exponential equation 2x+ 3 ·4x−1= 35.
Solution
Step 1: Rewrite the equation in terms of a single exponential base.
Let’s rewrite 4as 22, so our equation becomes:
2x+ 3 ·(22)x−1= 35.
Step 2: Simplify the equation using the properties of exponents.
Now we have:
2x+ 3 ·22x−2= 35.
Step 3: Use the property of exponents (am)n=amn.
Rewrite 22x−2as (22)x−1, which simplifies to 4x−1:
2x+ 3 ·4x−1= 35.
20
Step 4: Substitute 4x−1back into the equation.
The equation becomes:
2x+ 3 ·4x−1= 35.
Step 5: Now we have the original equation back.
2x+ 3 ·4x−1= 35.
Step 6: Solve the exponential equation.
Substitute 4x−1as 22·2x−1= 2x+1:
2x+ 3 ·2x+1 = 35.
Step 7: Combine the terms with the same base.
2x+ 3 ·2·2x= 35.
Step 8: Combine the terms.
2x+ 6 ·2x= 35.
Step 9: Simplify further.
7·2x= 35.
Step 10: Divide by 7to isolate 2x.
2x=35
7.
Step 11: Simplify the equation further.
2x= 5.
Step 12: Find the value of x.
Since 22= 4 and 23= 8, we can see that 2x= 5 when x= log25.
Therefore, the solution to the exponential equation is x= log25.
Question 31
Question
Solve the exponential equation 2x−1= 8.
21
Solution
Step 1: Rewrite 8as a power of 2.
23= 8
Step 2: Substitute 8as 23in the equation.
2x−1= 23
Step 3: Use the property of exponents that states if am=an, then m=n.
x−1 = 3
Step 4: Solve for x.
x= 3 + 1
x= 4
Therefore, the solution to the equation 2x−1= 8 is x= 4.
Question 32
Question
Solve the exponential equation 32x= 81.
Solution
Step 1: Rewrite 81 as a power of 3. Step 2: Simplify the equation by using the
properties of exponents. Step 3: Solve for x.
Step 1: Rewriting 81 as a power of 3:
81 = 34
Step 2: Simplifying the equation:
32x= 34
Since the bases are the same, we can set the exponents equal to each other:
2x= 4
Step 3: Solve for x: Dividing both sides by 2:
x= 2
Therefore, the solution to the exponential equation 32x= 81 is x= 2.
22
Question 33
Question
Solve the exponential equation: e2x= 10.
Solution
Step 1: Take the natural logarithm of both sides to isolate the variable:
ln(e2x)= ln(10)
Step 2: Use the property ln(ab)=bln(a)to simplify the left side:
2xln(e) = ln(10)
Step 3: Recall that ln(e) = 1, therefore we have:
2x= ln(10)
Step 4: Divide both sides by 2 to solve for x:
x=ln(10)
2
Step 5: Use a calculator to find the approximate value of x:
x≈ln(10)
2≈0.8686
So the solution to the equation e2x= 10 is x≈0.8686.
Question 34
Question
Solve the exponential equation 4x+ 32x= 125 for x.
Solution
Step 1: We can rewrite 125 as 53, so the equation becomes 4x+ 32x= 53.
Step 2: Rewrite 4as 22and 32xas (3x)2to get 22x+ (3x)2= 53.
Step 3: Rewrite 53as (2 ·5)3to make it easier to recognize a pattern.
Step 4: Rewrite the equation as 22x+ (3x)2= (2 ·5)3.
Step 5: Use the binomial formula to expand the right side: 22x+ (3x)2=
23·53.
Step 6: Simplify the right side to get 22x+ (3x)2= 8 ·125.
Step 7: Simplify further to obtain 22x+ (3x)2= 1000.
23
Step 8: Substitute a= 2xand b= 3xto get the quadratic equation a2+b2=
1000.
Step 9: Note that a= 10 and b= 30 is a solution, which implies 2x= 10
and 3x= 30.
Step 10: Solve 2x= 10 to find x= log210.
Step 11: Similarly, solve 3x= 30 to find x= log330.
Therefore, the solutions to the equation 4x+ 32x= 125 are x= log210 and
x= log330.
Question 35
Question
Solve the following exponential equation: 2x+3 = 16.
Solution
Step 1: Rewrite 16 as a power of 2:16 = 24.
Step 2: Substitute 16 as 24in the equation 2x+3 = 16 to get 2x+3 = 24.
Step 3: Since the bases are the same, we can set the exponents equal to each
other. So, we have x+ 3 = 4.
Step 4: Solve for xby subtracting 3from both sides: x= 1.
Therefore, the solution to the exponential equation 2x+3 = 16 is x= 1.
24
Question 2
Question
Solve the exponential equation 3x+1 −2·3x= 5.
Solution
Step 1: Let’s simplify the given exponential equation by factoring out 3x:
3x+1 −2·3x= 3x·31−2·3x
= 3 ·3x−2·3x
= (3 −2) ·3x
= 1 ·3x
= 3x
Step 2: After simplifying, our equation becomes 3x= 5.
Step 3: To solve 3x= 5, we can take the natural logarithm of both sides to
eliminate the exponential:
ln(3x) = ln(5)
Step 4: By the logarithmic property ln(ab)=bln(a), we have:
xln(3) = ln(5)
Step 5: Solve for xby dividing both sides by ln(3):
x=ln(5)
ln(3)
Therefore, the solution to the exponential equation 3x+1 −2·3x= 5 is
x=ln(5)
ln(3) .
Question 3
Question
Let f(x) = 3 ·2x+2. Find the equation of the asymptote for y= log2(f(x)).
Solution
Step 1: Rewrite f(x)using the properties of exponents.
f(x) = 3 ·2x+2
= 3 ·2x·22
= 12 ·2x
2
Step 2: Now, find log2(f(x)).
log2(f(x)) = log2(12 ·2x)
= log2(12) + log2(2x)
= log2(12) + x
Step 3: The equation of the asymptote for y= log2(f(x)) is y=x.
Question 4
Question
Suppose a certain investment grows according to the formula A(t) = P ert , where
A(t)represents the amount of money in the account after tyears, Pis the initial
investment, ris the annual interest rate (expressed as a decimal), and tis the
time in years.
A student invests $5000 in an account that offers 6How long will it take for
the investment to double in value?
Solution
Let’s first identify the given information: - Initial investment, P= $5000 -
Annual interest rate, r= 0.06 - Compounded continuously implies the formula
A(t) = P ert
We are looking for the time tit will take for the investment to double in
value. This means we are looking for the time tat which A(t) = 2P.
Step 1: Substitute the given values into the formula A(t) = P ert and set it
equal to 2P.
2P=P ert
Step 2: Cancel out the Pterms on both sides of the equation.
2 = ert
Step 3: Take the natural logarithm of both sides to solve for t.
ln 2 = ln(ert )
ln 2 = rt
Step 4: Divide both sides by rto solve for t.
t=ln 2
r
t=ln 2
0.06
3
Step 5: Calculate the value of t.
t≈0.693147
0.06
t≈11.55245
Therefore, it will take approximately 11.55 years for the investment to double
in value.
Question 5
Question
Samantha invested $10,000 in a savings account that yields an annual interest
rate of 4.5%, compounded continuously. How much money will Samantha have
in her account after 10 years?
Solution
Step 1: The formula for continuous compound interest is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount ($10,000 in this case), - ris the annual interest rate (4.5% or 0.045),
and - tis the time the money is invested for (10 years).
Step 2: Substitute the given values into the formula and solve for A:
A= 10000 ·e0.045·10
Step 3: Simplify the exponential expression:
A= 10000 ·e0.45
Step 4: Evaluate e0.45 using a calculator:
A≈10000 ·1.57029
Step 5: Calculate the final amount in Samantha’s account after 10 years:
A≈$15,702.90
Therefore, Samantha will have approximately $15,702.90 in her account after
10 years.
Question 6
Question
A population of bacteria in a petri dish is modeled by the function P(t) =
100e0.05t, where tis the time in hours. Find the rate of growth of the population
after 3 hours.
4
Solution
Step 1: To find the rate of growth of the population after 3 hours, we need to
find the derivative of the population function P(t)with respect to time t.
Step 2: Let P′(t)denote the rate of growth of the population at time t. We
have:
P′(t) = d
dt(100e0.05t)
Step 3: Using the chain rule of differentiation, we have:
P′(t) = 100 ·0.05e0.05t
Step 4: Simplifying the expression, we get:
P′(t) = 5e0.05t
Step 5: To find the rate of growth after 3 hours, we evaluate P′(3):
P′(3) = 5e0.05(3)
Step 6: Calculate the value:
P′(3) = 5e0.15 ≈5(1.161) ≈5.805
Step 7: Therefore, the rate of growth of the population after 3 hours is
approximately 5.805 bacteria per hour.
Question 7
Question
Samantha invests $10,000 in an account that earns 5% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Step 1: Express the continuous compound interest formula.
The continuous compound interest formula is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears - Pis the principal
amount (initial investment) - ris the annual interest rate (in decimal form) - e
is the base of the natural logarithm (approximately equal to 2.71828) - tis the
time the money is invested or borrowed for
Step 2: Set up the equation for doubling the initial investment.
When the initial investment doubles, the amount becomes 2P. Based on the
continuous compound interest formula, this occurs when:
2P=P·e0.05t
5
Step 3: Solve for the time tit takes for the investment to double.
Divide both sides of the equation by Pto isolate the exponential term:
2 = e0.05t
Step 4: Solve for tusing logarithms.
To solve for t, take the natural logarithm of both sides:
ln(2) = ln(e0.05t)
Step 5: Use the properties of logarithms to simplify.
The natural logarithm of the base eis just 1, so the equation simplifies to:
ln(2) = 0.05t
Step 6: Solve for t.
Divide by 0.05 to isolate t:
t=ln(2)
0.05
Step 7: Calculate the exact time it will take for the investment to double.
Substitute the values into the formula:
t=ln(2)
0.05 ≈0.6931
0.05 ≈13.86 years
Therefore, it will take approximately 13.86 years for Samantha’s investment
to double in value.
Question 8
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 can be written as 33.
Step 2: Set up the equation with the same base.
3x−1= 33.
Step 3: Equate the exponents.
x−1 = 3.
Step 4: Solve for x.
x= 3 + 1
6
x= 4.
Step 5: Check the solution by substituting x= 4 back into the original
equation.
34−1= 27
33= 27
27 = 27.
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 9
Question
Suppose a certain investment grows according to the exponential function A(t) =
5000 ·1.07t, where A(t)represents the amount of money in the account after t
years. Determine the time it will take for the initial investment of 5000totripleinvalue.
Solution
Step 1: Set up the equation for when the initial investment triples in value.
5000 ·1.07t= 3 ·5000
Step 2: Simplify the equation.
1.07t= 3
Step 3: Take the natural logarithm of both sides to solve for t. Recall that
if ex=y, then ln(y) = x. Applying this to our equation:
ln(1.07t)= ln(3)
tln(1.07) = ln(3)
Step 4: Solve for t.
t=ln(3)
ln(1.07)
Step 5: Using a calculator, calculate the value of t.
t≈ln(3)
ln(1.07) ≈1.09861
0.06303 ≈17.44
Step 6: Therefore, it will take approximately 17.44 years for the initial in-
vestment of 5000totripleinvalue.
7
Question 10
Question
Solve the exponential equation: 52x−1= 125.
Solution
Step 1: Rewrite both sides of the equation using the same base.
52x−1= 125
52x−1= 53
Step 2: Since the bases are the same, we can set the exponents equal to each
other.
2x−1 = 3
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 4: Check the solution.
52(2)−1= 125
53= 125
125 = 125
Therefore, the solution to the exponential equation 52x−1= 125 is x= 2.
Question 11
Question
Suppose a certain investment grows according to the exponential function A(t) =
5000 ×1.06twhere A(t)represents the amount of money invested after tyears.
(a) Find the initial investment.
(b) After how many years will the investment double?
(c) Determine the amount of money after 10 years if the investment is left
untouched.
8
Solution
(a) To find the initial investment, we look for the value of A(0). Substitute
t= 0 into the exponential function:
A(0) = 5000 ×1.060
A(0) = 5000 ×1
A(0) = 5000
Therefore, the initial investment was $5000.
(b) To find when the investment will double, we need to solve for tin the
equation A(t) = 2 ×A(0).
5000 ×1.06t= 2 ×5000
1.06t= 2
To solve for t, take the natural logarithm of both sides:
ln(1.06t)= ln(2)
tln(1.06) = ln(2)
t=ln(2)
ln(1.06) ≈11.89
Therefore, the investment will double after approximately 11.89 years.
(c) To determine the amount of money after 10 years, we can simply substi-
tute t= 10 into the exponential function:
A(10) = 5000 ×1.0610 ≈5000 ×1.790847 ≈8954.24
After 10 years, the amount of money in the investment will be approximately
$8954.24.
Question 12
Question
Samantha invests $5000 in a savings account that earns 3% annual interest
compounded quarterly. How much will Samantha have in her account after 10
years?
9
Solution
Step 1: First, we find the quarterly interest rate by dividing the annual interest
rate by 4. In this case, the quarterly interest rate is r=3%
4= 0.75% = 0.0075.
Step 2: We use the compound interest formula to determine the amount in
the account after 10 years:
A=P(1 + r
n)nt
where: - Ais the amount in the account after tyears, - Pis the principal amount
(initial investment), - ris the quarterly interest rate, - nis the number of times
interest is compounded per year, and - tis the number of years.
Step 3: Substituting the given values into the formula, we have:
A= 5000 (1 + 0.0075
4)4×10
Step 4: Simplifying, we get:
A= 5000 ×(1.001875)40
Step 5: Calculating the value inside the parentheses gives:
A= 5000 ×1.354035786
Step 6: Finally, we find:
A≈6770.18
Therefore, Samantha will have approximately $6770.18 in her account after
10 years.
Question 13
Question
Suppose a population of bacteria doubles every 3 hours. If there are initially
100 bacteria in the population, find an exponential model for the population
size after thours.
Solution
Step 1: Let’s denote the initial population of bacteria as P0= 100 and the
growth rate as r(since the population doubles every 3 hours, r= 2).
Step 2: The exponential growth formula is given by P(t) = P0·(1 + r)t
3,
where P(t)represents the population size at time t.
Step 3: Substitute P0= 100,r= 2, and simplify the equation to find the
exponential model for the population size after thours:
P(t) = 100 ·(1 + 2) t
3= 100 ·3t
3= 100 ·3t
3
Therefore, the exponential model for the population size after thours is
P(t) = 100 ·3t
3.
10
Question 14
Question
Let e2x·ex= 48. Find the exact value of x.
Solution
Step 1: Simplify the left side of the equation. We know that ea·eb=ea+b.
Therefore,
e2x·ex=e2x+x=e3x.
Step 2: Replace e3xin the original equation.
e3x= 48.
Step 3: Use the definition of the natural logarithm (ln) to solve for x. Take
the natural logarithm of both sides of the equation.
ln(e3x)= ln(48).
Step 4: Recall that ln(ea) = afor any real number a. Therefore, we have:
3x= ln(48).
Step 5: Finally, solve for xby dividing both sides by 3.
x=ln(48)
3.
So, the exact value of xis ln(48)
3.
Question 15
Question
The population of a town is modeled by the function P(t) = 2000 ·1.03twhere
trepresents time in years. Find the population after 10 years and the annual
growth rate of the population.
Solution
Step 1: To find the population after 10 years, plug in t= 10 into the population
function.
P(10) = 2000 ·1.0310
P(10) = 2000 ·1.344031
P(10) = 2688.062
So, the population after 10 years is 2688.062.
Step 2: The annual growth rate, which is the factor by which the population
increases each year, is given by the multiplier in the exponential function. In
this case, the growth rate is 1.03 or 3
Therefore, the population grows at an annual rate of 3
11
Question 16
Question
Suppose the population of a town is modeled by the function P(t) = 5000·1.03t,
where trepresents time in years.
How long will it take for the population of the town to double?
Solution
Step 1: Set up the equation for population doubling.
We need to find the time tit takes for the population to double, which means
P(t) = 2 ·P(0), where P(0) is the initial population.
2·5000 = 5000 ·1.03t
Step 2: Solve for t.
Divide both sides by 5000:
2 = 1.03t
Step 3: Take the natural logarithm of both sides.
ln(2) = ln(1.03t)
Step 4: Use the property of logarithms ln(ab)=b·ln(a)to simplify the right
side.
ln(2) = t·ln(1.03)
Step 5: Solve for tby dividing both sides by ln(1.03).
t=ln(2)
ln(1.03)
Step 6: Calculate the approximate value of t.
t≈0.6931
0.02956 ≈23.44
Therefore, it will take approximately 23.44 years for the population of the
town to double.
Question 17
Question
Solve the exponential equation 23x= 16.
12
Solution
Step 1: Rewrite 16 as a power of 2.
16 = 24
Step 2: Substitute 16 as 24in the equation 23x= 16.
23x= 24
Step 3: Equate the exponents.
3x= 4
Step 4: Solve for xby dividing both sides by 3.
x=4
3
Therefore, the solution to the equation 23x= 16 is x=4
3.
Question 18
Question
Solve the exponential equation 42x−1= 64.
Solution
Step 1: Rewrite 4and 64 in terms of the same base. Step 2: Solve the resulting
exponential equation for x.
Step 1: First, rewrite both 4and 64 with the base 4:
42x−1= 64
(22)2x−1= 43
22(2x−1) = 26
24x−2= 26
Step 2: Since the bases are the same, we can equate the exponents:
4x−2 = 6
4x= 8
x= 2
Therefore, the solution to the equation 42x−1= 64 is x= 2.
13
Question 19
Question
Solve the equation 32x= 125 for x.
Solution
Step 1: Rewrite 125 as a power of 3. Since 125 = 33, the equation becomes
32x= 33.
Step 2: Set the exponents equal to each other: 2x= 3.
Step 3: Solve for xby dividing both sides by 2: x=3
2.
Therefore, the solution to the equation 32x= 125 is x=3
2.
Question 20
Question
Sara invests $5,000 in an account that earns 8% interest compounded annually.
How much money will be in the account after 10 years? Round your answer to
the nearest dollar.
Solution
Step 1: First, we will use the formula for compound interest to find the future
value of the investment:
A=P(1 + r
n)nt
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment) of $5,000, - ris the annual interest rate of 8% or
0.08, - nis the number of times the interest is compounded per year (in this
case, once annually), - tis the number of years the money is invested for.
Step 2: Substitute the given values into the formula:
A= 5000 (1 + 0.08
1)1∗10
Step 3: Calculate the future value of the investment:
A= 5000 ×(1 + 0.08)10
A= 5000 ×(1.08)10
Step 4: Calculate (1.08)10:
(1.08)10 ≈2.158924
14
Step 5: Substitute the approximation back into the equation:
A= 5000 ×2.158924
A≈10794.62
Step 6: Round the answer to the nearest dollar: After 10 years, the amount
of money in the account will be approximately $10,795.
Question 21
Question
Solve the following exponential equation for x:32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the two exponential expressions equal to each other.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
Step 4: Solve for x.
2x= 2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation:
32(1)+1 = 27
33= 27
27 = 27
Step 6: Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 22
Question
A certain investment grows exponentially. If the initial investment was $10,000
and it doubled in value after 10 years, determine the growth rate of the invest-
ment.
15
Solution
Let Pbe the initial investment, rbe the growth rate, and tbe the time in years.
Step 1: Use the formula for exponential growth to set up the equation.
Given that the investment doubles in value after 10 years, we have:
2P=P×ert
Step 2: Plug in the given values. We know that the initial investment
P= $10,000 and the time t= 10 years. Substituting these values into the
equation, we get:
2(10000) = 10000 ×e10r
Step 3: Solve for the growth rate r. Dividing both sides by 10000, we get:
2 = e10r
Taking the natural logarithm of both sides to solve for r, we get:
ln(2) = ln(e10r)
ln(2) = 10r
r=ln(2)
10
Step 4: Calculate the growth rate r. Using a calculator, we find
r≈ln(2)
10 ≈0.0693
Therefore, the growth rate of the investment is approximately 0.0693, or
6.93
Question 23
Question
Solve the exponential equation: 3x+1 −3x= 20.
Solution
Step 1: Use the properties of exponents to simplify the equation. Step 2: Let
u= 3x. Step 3: Substitute uinto the equation. Step 4: Solve for u. Step 5:
Substitute back to solve for x.
Step 1: Simplify the equation using the properties of exponents:
3x+1 −3x= 20
3·3x−3x= 20
16
3·3x−3x= 20
Step 2: Let u= 3x. Step 3: Substitute uinto the equation:
3u−u= 20
2u= 20
Step 4: Solve for u:
u=20
2
u= 10
Step 5: Substitute back to solve for x:
3x= 10
x= log3(10)
Question 24
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
We know that 27 = 33, so the equation becomes 3x−1= 33.
Step 2: Set the exponents equal to each other.
Since the bases are the same, we can set the exponents equal: x−1 = 3.
Step 3: Solve for x.
Add 1to both sides to solve for x:x= 4.
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 25
Question
Solve the exponential equation 32x−3= 27.
Solution
Step 1: Rewrite 27 using base 3. Step 2: Solve for x.
17
Question 26
Question
Solve the exponential equation 2·32x−1= 6x+ 18 for x.
Solution
Step 1: Rewrite both sides with the same base. Step 2: Use the properties of
exponents to simplify the equation. Step 3: Solve the resulting equation for x.
Step 1: Rewrite both sides with the same base.
We can rewrite 6as 32, so the given equation becomes:
2·32x−1= (32)x+ 18
Step 2: Use the properties of exponents to simplify the equation.
Now, simplify the right side:
2·32x−1= 32x+ 18
Step 3: Solve the resulting equation for x.
So, the equation we need to solve is:
2·32x−1= 32x+ 18
Let’s now substitute 32xas afor simplification:
2·32x−1=a+ 18
2·32x−1=a+ 18
2·32x−1=a+ 18
So, the equation to solve is:
2·32x−1=a+ 18
Can you figure out how to solve it now?
Question 27
Question
The population of a town is modeled by the function P(t) = 5000×1.03t, where
trepresents the number of years since the year 2020.
27. Determine the population of the town in the year 2030 according to the
model.
18
Solution
Step 1: First, we need to determine the value of tfor the year 2030. Since t
represents the number of years since 2020, in the year 2030, t= 2030−2020 = 10.
Step 2: Substitute t= 10 into the function P(t)to find the population in
the year 2030.
P(10) = 5000 ×1.0310
Step 3: Calculate 1.0310:
1.0310 ≈1.344031
Step 4: Multiply 5000 by 1.344031 to find the population in the year 2030:
P(10) ≈5000 ×1.344031 ≈6720.155
Therefore, according to the model, the population of the town in the year
2030 is approximately 6720.
Question 28
Question
Solve the following exponential equation:
5x−1= 125
Solution
Step 1: Rewrite 125 as a power of 5:
125 = 53
Step 2: Substitute the expression for 125 into the equation:
5x−1= 53
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
x−1 = 3
Step 4: Solve for x:
x= 3 + 1
x= 4
Step 5: Therefore, the solution to the exponential equation is x= 4.
19
Question 29
Question
Solve the following exponential equation for x:32x= 81.
Solution
Step 1: Rewrite 81 as a power of 3.
81 = 34
Step 2: Set the two expressions equal to each other.
32x= 34
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x= 4
Step 4: Solve for x.
x=4
2
x= 2
Step 5: Therefore, the solution to the equation 32x= 81 is x= 2.
Question 30
Question
Solve the exponential equation 2x+ 3 ·4x−1= 35.
Solution
Step 1: Rewrite the equation in terms of a single exponential base.
Let’s rewrite 4as 22, so our equation becomes:
2x+ 3 ·(22)x−1= 35.
Step 2: Simplify the equation using the properties of exponents.
Now we have:
2x+ 3 ·22x−2= 35.
Step 3: Use the property of exponents (am)n=amn.
Rewrite 22x−2as (22)x−1, which simplifies to 4x−1:
2x+ 3 ·4x−1= 35.
20
Step 4: Substitute 4x−1back into the equation.
The equation becomes:
2x+ 3 ·4x−1= 35.
Step 5: Now we have the original equation back.
2x+ 3 ·4x−1= 35.
Step 6: Solve the exponential equation.
Substitute 4x−1as 22·2x−1= 2x+1:
2x+ 3 ·2x+1 = 35.
Step 7: Combine the terms with the same base.
2x+ 3 ·2·2x= 35.
Step 8: Combine the terms.
2x+ 6 ·2x= 35.
Step 9: Simplify further.
7·2x= 35.
Step 10: Divide by 7to isolate 2x.
2x=35
7.
Step 11: Simplify the equation further.
2x= 5.
Step 12: Find the value of x.
Since 22= 4 and 23= 8, we can see that 2x= 5 when x= log25.
Therefore, the solution to the exponential equation is x= log25.
Question 31
Question
Solve the exponential equation 2x−1= 8.
21
Solution
Step 1: Rewrite 8as a power of 2.
23= 8
Step 2: Substitute 8as 23in the equation.
2x−1= 23
Step 3: Use the property of exponents that states if am=an, then m=n.
x−1 = 3
Step 4: Solve for x.
x= 3 + 1
x= 4
Therefore, the solution to the equation 2x−1= 8 is x= 4.
Question 32
Question
Solve the exponential equation 32x= 81.
Solution
Step 1: Rewrite 81 as a power of 3. Step 2: Simplify the equation by using the
properties of exponents. Step 3: Solve for x.
Step 1: Rewriting 81 as a power of 3:
81 = 34
Step 2: Simplifying the equation:
32x= 34
Since the bases are the same, we can set the exponents equal to each other:
2x= 4
Step 3: Solve for x: Dividing both sides by 2:
x= 2
Therefore, the solution to the exponential equation 32x= 81 is x= 2.
22
Question 33
Question
Solve the exponential equation: e2x= 10.
Solution
Step 1: Take the natural logarithm of both sides to isolate the variable:
ln(e2x)= ln(10)
Step 2: Use the property ln(ab)=bln(a)to simplify the left side:
2xln(e) = ln(10)
Step 3: Recall that ln(e) = 1, therefore we have:
2x= ln(10)
Step 4: Divide both sides by 2 to solve for x:
x=ln(10)
2
Step 5: Use a calculator to find the approximate value of x:
x≈ln(10)
2≈0.8686
So the solution to the equation e2x= 10 is x≈0.8686.
Question 34
Question
Solve the exponential equation 4x+ 32x= 125 for x.
Solution
Step 1: We can rewrite 125 as 53, so the equation becomes 4x+ 32x= 53.
Step 2: Rewrite 4as 22and 32xas (3x)2to get 22x+ (3x)2= 53.
Step 3: Rewrite 53as (2 ·5)3to make it easier to recognize a pattern.
Step 4: Rewrite the equation as 22x+ (3x)2= (2 ·5)3.
Step 5: Use the binomial formula to expand the right side: 22x+ (3x)2=
23·53.
Step 6: Simplify the right side to get 22x+ (3x)2= 8 ·125.
Step 7: Simplify further to obtain 22x+ (3x)2= 1000.
23
Step 8: Substitute a= 2xand b= 3xto get the quadratic equation a2+b2=
1000.
Step 9: Note that a= 10 and b= 30 is a solution, which implies 2x= 10
and 3x= 30.
Step 10: Solve 2x= 10 to find x= log210.
Step 11: Similarly, solve 3x= 30 to find x= log330.
Therefore, the solutions to the equation 4x+ 32x= 125 are x= log210 and
x= log330.
Question 35
Question
Solve the following exponential equation: 2x+3 = 16.
Solution
Step 1: Rewrite 16 as a power of 2:16 = 24.
Step 2: Substitute 16 as 24in the equation 2x+3 = 16 to get 2x+3 = 24.
Step 3: Since the bases are the same, we can set the exponents equal to each
other. So, we have x+ 3 = 4.
Step 4: Solve for xby subtracting 3from both sides: x= 1.
Therefore, the solution to the exponential equation 2x+3 = 16 is x= 1.
24
Question 2
Question
Solve the exponential equation 3x+1 −2·3x= 5.
Solution
Step 1: Let’s simplify the given exponential equation by factoring out 3x:
3x+1 −2·3x= 3x·31−2·3x
= 3 ·3x−2·3x
= (3 −2) ·3x
= 1 ·3x
= 3x
Step 2: After simplifying, our equation becomes 3x= 5.
Step 3: To solve 3x= 5, we can take the natural logarithm of both sides to
eliminate the exponential:
ln(3x) = ln(5)
Step 4: By the logarithmic property ln(ab)=bln(a), we have:
xln(3) = ln(5)
Step 5: Solve for xby dividing both sides by ln(3):
x=ln(5)
ln(3)
Therefore, the solution to the exponential equation 3x+1 −2·3x= 5 is
x=ln(5)
ln(3) .
Question 3
Question
Let f(x) = 3 ·2x+2. Find the equation of the asymptote for y= log2(f(x)).
Solution
Step 1: Rewrite f(x)using the properties of exponents.
f(x) = 3 ·2x+2
= 3 ·2x·22
= 12 ·2x
2
Step 2: Now, find log2(f(x)).
log2(f(x)) = log2(12 ·2x)
= log2(12) + log2(2x)
= log2(12) + x
Step 3: The equation of the asymptote for y= log2(f(x)) is y=x.
Question 4
Question
Suppose a certain investment grows according to the formula A(t) = P ert , where
A(t)represents the amount of money in the account after tyears, Pis the initial
investment, ris the annual interest rate (expressed as a decimal), and tis the
time in years.
A student invests $5000 in an account that offers 6How long will it take for
the investment to double in value?
Solution
Let’s first identify the given information: - Initial investment, P= $5000 -
Annual interest rate, r= 0.06 - Compounded continuously implies the formula
A(t) = P ert
We are looking for the time tit will take for the investment to double in
value. This means we are looking for the time tat which A(t) = 2P.
Step 1: Substitute the given values into the formula A(t) = P ert and set it
equal to 2P.
2P=P ert
Step 2: Cancel out the Pterms on both sides of the equation.
2 = ert
Step 3: Take the natural logarithm of both sides to solve for t.
ln 2 = ln(ert )
ln 2 = rt
Step 4: Divide both sides by rto solve for t.
t=ln 2
r
t=ln 2
0.06
3
Step 5: Calculate the value of t.
t≈0.693147
0.06
t≈11.55245
Therefore, it will take approximately 11.55 years for the investment to double
in value.
Question 5
Question
Samantha invested $10,000 in a savings account that yields an annual interest
rate of 4.5%, compounded continuously. How much money will Samantha have
in her account after 10 years?
Solution
Step 1: The formula for continuous compound interest is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount ($10,000 in this case), - ris the annual interest rate (4.5% or 0.045),
and - tis the time the money is invested for (10 years).
Step 2: Substitute the given values into the formula and solve for A:
A= 10000 ·e0.045·10
Step 3: Simplify the exponential expression:
A= 10000 ·e0.45
Step 4: Evaluate e0.45 using a calculator:
A≈10000 ·1.57029
Step 5: Calculate the final amount in Samantha’s account after 10 years:
A≈$15,702.90
Therefore, Samantha will have approximately $15,702.90 in her account after
10 years.
Question 6
Question
A population of bacteria in a petri dish is modeled by the function P(t) =
100e0.05t, where tis the time in hours. Find the rate of growth of the population
after 3 hours.
4
Solution
Step 1: To find the rate of growth of the population after 3 hours, we need to
find the derivative of the population function P(t)with respect to time t.
Step 2: Let P′(t)denote the rate of growth of the population at time t. We
have:
P′(t) = d
dt(100e0.05t)
Step 3: Using the chain rule of differentiation, we have:
P′(t) = 100 ·0.05e0.05t
Step 4: Simplifying the expression, we get:
P′(t) = 5e0.05t
Step 5: To find the rate of growth after 3 hours, we evaluate P′(3):
P′(3) = 5e0.05(3)
Step 6: Calculate the value:
P′(3) = 5e0.15 ≈5(1.161) ≈5.805
Step 7: Therefore, the rate of growth of the population after 3 hours is
approximately 5.805 bacteria per hour.
Question 7
Question
Samantha invests $10,000 in an account that earns 5% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Step 1: Express the continuous compound interest formula.
The continuous compound interest formula is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears - Pis the principal
amount (initial investment) - ris the annual interest rate (in decimal form) - e
is the base of the natural logarithm (approximately equal to 2.71828) - tis the
time the money is invested or borrowed for
Step 2: Set up the equation for doubling the initial investment.
When the initial investment doubles, the amount becomes 2P. Based on the
continuous compound interest formula, this occurs when:
2P=P·e0.05t
5
Step 3: Solve for the time tit takes for the investment to double.
Divide both sides of the equation by Pto isolate the exponential term:
2 = e0.05t
Step 4: Solve for tusing logarithms.
To solve for t, take the natural logarithm of both sides:
ln(2) = ln(e0.05t)
Step 5: Use the properties of logarithms to simplify.
The natural logarithm of the base eis just 1, so the equation simplifies to:
ln(2) = 0.05t
Step 6: Solve for t.
Divide by 0.05 to isolate t:
t=ln(2)
0.05
Step 7: Calculate the exact time it will take for the investment to double.
Substitute the values into the formula:
t=ln(2)
0.05 ≈0.6931
0.05 ≈13.86 years
Therefore, it will take approximately 13.86 years for Samantha’s investment
to double in value.
Question 8
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 can be written as 33.
Step 2: Set up the equation with the same base.
3x−1= 33.
Step 3: Equate the exponents.
x−1 = 3.
Step 4: Solve for x.
x= 3 + 1
6
x= 4.
Step 5: Check the solution by substituting x= 4 back into the original
equation.
34−1= 27
33= 27
27 = 27.
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 9
Question
Suppose a certain investment grows according to the exponential function A(t) =
5000 ·1.07t, where A(t)represents the amount of money in the account after t
years. Determine the time it will take for the initial investment of 5000totripleinvalue.
Solution
Step 1: Set up the equation for when the initial investment triples in value.
5000 ·1.07t= 3 ·5000
Step 2: Simplify the equation.
1.07t= 3
Step 3: Take the natural logarithm of both sides to solve for t. Recall that
if ex=y, then ln(y) = x. Applying this to our equation:
ln(1.07t)= ln(3)
tln(1.07) = ln(3)
Step 4: Solve for t.
t=ln(3)
ln(1.07)
Step 5: Using a calculator, calculate the value of t.
t≈ln(3)
ln(1.07) ≈1.09861
0.06303 ≈17.44
Step 6: Therefore, it will take approximately 17.44 years for the initial in-
vestment of 5000totripleinvalue.
7
Question 10
Question
Solve the exponential equation: 52x−1= 125.
Solution
Step 1: Rewrite both sides of the equation using the same base.
52x−1= 125
52x−1= 53
Step 2: Since the bases are the same, we can set the exponents equal to each
other.
2x−1 = 3
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 4: Check the solution.
52(2)−1= 125
53= 125
125 = 125
Therefore, the solution to the exponential equation 52x−1= 125 is x= 2.
Question 11
Question
Suppose a certain investment grows according to the exponential function A(t) =
5000 ×1.06twhere A(t)represents the amount of money invested after tyears.
(a) Find the initial investment.
(b) After how many years will the investment double?
(c) Determine the amount of money after 10 years if the investment is left
untouched.
8
Solution
(a) To find the initial investment, we look for the value of A(0). Substitute
t= 0 into the exponential function:
A(0) = 5000 ×1.060
A(0) = 5000 ×1
A(0) = 5000
Therefore, the initial investment was $5000.
(b) To find when the investment will double, we need to solve for tin the
equation A(t) = 2 ×A(0).
5000 ×1.06t= 2 ×5000
1.06t= 2
To solve for t, take the natural logarithm of both sides:
ln(1.06t)= ln(2)
tln(1.06) = ln(2)
t=ln(2)
ln(1.06) ≈11.89
Therefore, the investment will double after approximately 11.89 years.
(c) To determine the amount of money after 10 years, we can simply substi-
tute t= 10 into the exponential function:
A(10) = 5000 ×1.0610 ≈5000 ×1.790847 ≈8954.24
After 10 years, the amount of money in the investment will be approximately
$8954.24.
Question 12
Question
Samantha invests $5000 in a savings account that earns 3% annual interest
compounded quarterly. How much will Samantha have in her account after 10
years?
9
Solution
Step 1: First, we find the quarterly interest rate by dividing the annual interest
rate by 4. In this case, the quarterly interest rate is r=3%
4= 0.75% = 0.0075.
Step 2: We use the compound interest formula to determine the amount in
the account after 10 years:
A=P(1 + r
n)nt
where: - Ais the amount in the account after tyears, - Pis the principal amount
(initial investment), - ris the quarterly interest rate, - nis the number of times
interest is compounded per year, and - tis the number of years.
Step 3: Substituting the given values into the formula, we have:
A= 5000 (1 + 0.0075
4)4×10
Step 4: Simplifying, we get:
A= 5000 ×(1.001875)40
Step 5: Calculating the value inside the parentheses gives:
A= 5000 ×1.354035786
Step 6: Finally, we find:
A≈6770.18
Therefore, Samantha will have approximately $6770.18 in her account after
10 years.
Question 13
Question
Suppose a population of bacteria doubles every 3 hours. If there are initially
100 bacteria in the population, find an exponential model for the population
size after thours.
Solution
Step 1: Let’s denote the initial population of bacteria as P0= 100 and the
growth rate as r(since the population doubles every 3 hours, r= 2).
Step 2: The exponential growth formula is given by P(t) = P0·(1 + r)t
3,
where P(t)represents the population size at time t.
Step 3: Substitute P0= 100,r= 2, and simplify the equation to find the
exponential model for the population size after thours:
P(t) = 100 ·(1 + 2) t
3= 100 ·3t
3= 100 ·3t
3
Therefore, the exponential model for the population size after thours is
P(t) = 100 ·3t
3.
10
Question 14
Question
Let e2x·ex= 48. Find the exact value of x.
Solution
Step 1: Simplify the left side of the equation. We know that ea·eb=ea+b.
Therefore,
e2x·ex=e2x+x=e3x.
Step 2: Replace e3xin the original equation.
e3x= 48.
Step 3: Use the definition of the natural logarithm (ln) to solve for x. Take
the natural logarithm of both sides of the equation.
ln(e3x)= ln(48).
Step 4: Recall that ln(ea) = afor any real number a. Therefore, we have:
3x= ln(48).
Step 5: Finally, solve for xby dividing both sides by 3.
x=ln(48)
3.
So, the exact value of xis ln(48)
3.
Question 15
Question
The population of a town is modeled by the function P(t) = 2000 ·1.03twhere
trepresents time in years. Find the population after 10 years and the annual
growth rate of the population.
Solution
Step 1: To find the population after 10 years, plug in t= 10 into the population
function.
P(10) = 2000 ·1.0310
P(10) = 2000 ·1.344031
P(10) = 2688.062
So, the population after 10 years is 2688.062.
Step 2: The annual growth rate, which is the factor by which the population
increases each year, is given by the multiplier in the exponential function. In
this case, the growth rate is 1.03 or 3
Therefore, the population grows at an annual rate of 3
11
Question 16
Question
Suppose the population of a town is modeled by the function P(t) = 5000·1.03t,
where trepresents time in years.
How long will it take for the population of the town to double?
Solution
Step 1: Set up the equation for population doubling.
We need to find the time tit takes for the population to double, which means
P(t) = 2 ·P(0), where P(0) is the initial population.
2·5000 = 5000 ·1.03t
Step 2: Solve for t.
Divide both sides by 5000:
2 = 1.03t
Step 3: Take the natural logarithm of both sides.
ln(2) = ln(1.03t)
Step 4: Use the property of logarithms ln(ab)=b·ln(a)to simplify the right
side.
ln(2) = t·ln(1.03)
Step 5: Solve for tby dividing both sides by ln(1.03).
t=ln(2)
ln(1.03)
Step 6: Calculate the approximate value of t.
t≈0.6931
0.02956 ≈23.44
Therefore, it will take approximately 23.44 years for the population of the
town to double.
Question 17
Question
Solve the exponential equation 23x= 16.
12
Solution
Step 1: Rewrite 16 as a power of 2.
16 = 24
Step 2: Substitute 16 as 24in the equation 23x= 16.
23x= 24
Step 3: Equate the exponents.
3x= 4
Step 4: Solve for xby dividing both sides by 3.
x=4
3
Therefore, the solution to the equation 23x= 16 is x=4
3.
Question 18
Question
Solve the exponential equation 42x−1= 64.
Solution
Step 1: Rewrite 4and 64 in terms of the same base. Step 2: Solve the resulting
exponential equation for x.
Step 1: First, rewrite both 4and 64 with the base 4:
42x−1= 64
(22)2x−1= 43
22(2x−1) = 26
24x−2= 26
Step 2: Since the bases are the same, we can equate the exponents:
4x−2 = 6
4x= 8
x= 2
Therefore, the solution to the equation 42x−1= 64 is x= 2.
13
Question 19
Question
Solve the equation 32x= 125 for x.
Solution
Step 1: Rewrite 125 as a power of 3. Since 125 = 33, the equation becomes
32x= 33.
Step 2: Set the exponents equal to each other: 2x= 3.
Step 3: Solve for xby dividing both sides by 2: x=3
2.
Therefore, the solution to the equation 32x= 125 is x=3
2.
Question 20
Question
Sara invests $5,000 in an account that earns 8% interest compounded annually.
How much money will be in the account after 10 years? Round your answer to
the nearest dollar.
Solution
Step 1: First, we will use the formula for compound interest to find the future
value of the investment:
A=P(1 + r
n)nt
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment) of $5,000, - ris the annual interest rate of 8% or
0.08, - nis the number of times the interest is compounded per year (in this
case, once annually), - tis the number of years the money is invested for.
Step 2: Substitute the given values into the formula:
A= 5000 (1 + 0.08
1)1∗10
Step 3: Calculate the future value of the investment:
A= 5000 ×(1 + 0.08)10
A= 5000 ×(1.08)10
Step 4: Calculate (1.08)10:
(1.08)10 ≈2.158924
14
Step 5: Substitute the approximation back into the equation:
A= 5000 ×2.158924
A≈10794.62
Step 6: Round the answer to the nearest dollar: After 10 years, the amount
of money in the account will be approximately $10,795.
Question 21
Question
Solve the following exponential equation for x:32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the two exponential expressions equal to each other.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
Step 4: Solve for x.
2x= 2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation:
32(1)+1 = 27
33= 27
27 = 27
Step 6: Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 22
Question
A certain investment grows exponentially. If the initial investment was $10,000
and it doubled in value after 10 years, determine the growth rate of the invest-
ment.
15
Solution
Let Pbe the initial investment, rbe the growth rate, and tbe the time in years.
Step 1: Use the formula for exponential growth to set up the equation.
Given that the investment doubles in value after 10 years, we have:
2P=P×ert
Step 2: Plug in the given values. We know that the initial investment
P= $10,000 and the time t= 10 years. Substituting these values into the
equation, we get:
2(10000) = 10000 ×e10r
Step 3: Solve for the growth rate r. Dividing both sides by 10000, we get:
2 = e10r
Taking the natural logarithm of both sides to solve for r, we get:
ln(2) = ln(e10r)
ln(2) = 10r
r=ln(2)
10
Step 4: Calculate the growth rate r. Using a calculator, we find
r≈ln(2)
10 ≈0.0693
Therefore, the growth rate of the investment is approximately 0.0693, or
6.93
Question 23
Question
Solve the exponential equation: 3x+1 −3x= 20.
Solution
Step 1: Use the properties of exponents to simplify the equation. Step 2: Let
u= 3x. Step 3: Substitute uinto the equation. Step 4: Solve for u. Step 5:
Substitute back to solve for x.
Step 1: Simplify the equation using the properties of exponents:
3x+1 −3x= 20
3·3x−3x= 20
16
3·3x−3x= 20
Step 2: Let u= 3x. Step 3: Substitute uinto the equation:
3u−u= 20
2u= 20
Step 4: Solve for u:
u=20
2
u= 10
Step 5: Substitute back to solve for x:
3x= 10
x= log3(10)
Question 24
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
We know that 27 = 33, so the equation becomes 3x−1= 33.
Step 2: Set the exponents equal to each other.
Since the bases are the same, we can set the exponents equal: x−1 = 3.
Step 3: Solve for x.
Add 1to both sides to solve for x:x= 4.
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 25
Question
Solve the exponential equation 32x−3= 27.
Solution
Step 1: Rewrite 27 using base 3. Step 2: Solve for x.
17
Question 26
Question
Solve the exponential equation 2·32x−1= 6x+ 18 for x.
Solution
Step 1: Rewrite both sides with the same base. Step 2: Use the properties of
exponents to simplify the equation. Step 3: Solve the resulting equation for x.
Step 1: Rewrite both sides with the same base.
We can rewrite 6as 32, so the given equation becomes:
2·32x−1= (32)x+ 18
Step 2: Use the properties of exponents to simplify the equation.
Now, simplify the right side:
2·32x−1= 32x+ 18
Step 3: Solve the resulting equation for x.
So, the equation we need to solve is:
2·32x−1= 32x+ 18
Let’s now substitute 32xas afor simplification:
2·32x−1=a+ 18
2·32x−1=a+ 18
2·32x−1=a+ 18
So, the equation to solve is:
2·32x−1=a+ 18
Can you figure out how to solve it now?
Question 27
Question
The population of a town is modeled by the function P(t) = 5000×1.03t, where
trepresents the number of years since the year 2020.
27. Determine the population of the town in the year 2030 according to the
model.
18
Solution
Step 1: First, we need to determine the value of tfor the year 2030. Since t
represents the number of years since 2020, in the year 2030, t= 2030−2020 = 10.
Step 2: Substitute t= 10 into the function P(t)to find the population in
the year 2030.
P(10) = 5000 ×1.0310
Step 3: Calculate 1.0310:
1.0310 ≈1.344031
Step 4: Multiply 5000 by 1.344031 to find the population in the year 2030:
P(10) ≈5000 ×1.344031 ≈6720.155
Therefore, according to the model, the population of the town in the year
2030 is approximately 6720.
Question 28
Question
Solve the following exponential equation:
5x−1= 125
Solution
Step 1: Rewrite 125 as a power of 5:
125 = 53
Step 2: Substitute the expression for 125 into the equation:
5x−1= 53
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
x−1 = 3
Step 4: Solve for x:
x= 3 + 1
x= 4
Step 5: Therefore, the solution to the exponential equation is x= 4.
19
Question 29
Question
Solve the following exponential equation for x:32x= 81.
Solution
Step 1: Rewrite 81 as a power of 3.
81 = 34
Step 2: Set the two expressions equal to each other.
32x= 34
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x= 4
Step 4: Solve for x.
x=4
2
x= 2
Step 5: Therefore, the solution to the equation 32x= 81 is x= 2.
Question 30
Question
Solve the exponential equation 2x+ 3 ·4x−1= 35.
Solution
Step 1: Rewrite the equation in terms of a single exponential base.
Let’s rewrite 4as 22, so our equation becomes:
2x+ 3 ·(22)x−1= 35.
Step 2: Simplify the equation using the properties of exponents.
Now we have:
2x+ 3 ·22x−2= 35.
Step 3: Use the property of exponents (am)n=amn.
Rewrite 22x−2as (22)x−1, which simplifies to 4x−1:
2x+ 3 ·4x−1= 35.
20
Step 4: Substitute 4x−1back into the equation.
The equation becomes:
2x+ 3 ·4x−1= 35.
Step 5: Now we have the original equation back.
2x+ 3 ·4x−1= 35.
Step 6: Solve the exponential equation.
Substitute 4x−1as 22·2x−1= 2x+1:
2x+ 3 ·2x+1 = 35.
Step 7: Combine the terms with the same base.
2x+ 3 ·2·2x= 35.
Step 8: Combine the terms.
2x+ 6 ·2x= 35.
Step 9: Simplify further.
7·2x= 35.
Step 10: Divide by 7to isolate 2x.
2x=35
7.
Step 11: Simplify the equation further.
2x= 5.
Step 12: Find the value of x.
Since 22= 4 and 23= 8, we can see that 2x= 5 when x= log25.
Therefore, the solution to the exponential equation is x= log25.
Question 31
Question
Solve the exponential equation 2x−1= 8.
21
Solution
Step 1: Rewrite 8as a power of 2.
23= 8
Step 2: Substitute 8as 23in the equation.
2x−1= 23
Step 3: Use the property of exponents that states if am=an, then m=n.
x−1 = 3
Step 4: Solve for x.
x= 3 + 1
x= 4
Therefore, the solution to the equation 2x−1= 8 is x= 4.
Question 32
Question
Solve the exponential equation 32x= 81.
Solution
Step 1: Rewrite 81 as a power of 3. Step 2: Simplify the equation by using the
properties of exponents. Step 3: Solve for x.
Step 1: Rewriting 81 as a power of 3:
81 = 34
Step 2: Simplifying the equation:
32x= 34
Since the bases are the same, we can set the exponents equal to each other:
2x= 4
Step 3: Solve for x: Dividing both sides by 2:
x= 2
Therefore, the solution to the exponential equation 32x= 81 is x= 2.
22
Question 33
Question
Solve the exponential equation: e2x= 10.
Solution
Step 1: Take the natural logarithm of both sides to isolate the variable:
ln(e2x)= ln(10)
Step 2: Use the property ln(ab)=bln(a)to simplify the left side:
2xln(e) = ln(10)
Step 3: Recall that ln(e) = 1, therefore we have:
2x= ln(10)
Step 4: Divide both sides by 2 to solve for x:
x=ln(10)
2
Step 5: Use a calculator to find the approximate value of x:
x≈ln(10)
2≈0.8686
So the solution to the equation e2x= 10 is x≈0.8686.
Question 34
Question
Solve the exponential equation 4x+ 32x= 125 for x.
Solution
Step 1: We can rewrite 125 as 53, so the equation becomes 4x+ 32x= 53.
Step 2: Rewrite 4as 22and 32xas (3x)2to get 22x+ (3x)2= 53.
Step 3: Rewrite 53as (2 ·5)3to make it easier to recognize a pattern.
Step 4: Rewrite the equation as 22x+ (3x)2= (2 ·5)3.
Step 5: Use the binomial formula to expand the right side: 22x+ (3x)2=
23·53.
Step 6: Simplify the right side to get 22x+ (3x)2= 8 ·125.
Step 7: Simplify further to obtain 22x+ (3x)2= 1000.
23
Step 8: Substitute a= 2xand b= 3xto get the quadratic equation a2+b2=
1000.
Step 9: Note that a= 10 and b= 30 is a solution, which implies 2x= 10
and 3x= 30.
Step 10: Solve 2x= 10 to find x= log210.
Step 11: Similarly, solve 3x= 30 to find x= log330.
Therefore, the solutions to the equation 4x+ 32x= 125 are x= log210 and
x= log330.
Question 35
Question
Solve the following exponential equation: 2x+3 = 16.
Solution
Step 1: Rewrite 16 as a power of 2:16 = 24.
Step 2: Substitute 16 as 24in the equation 2x+3 = 16 to get 2x+3 = 24.
Step 3: Since the bases are the same, we can set the exponents equal to each
other. So, we have x+ 3 = 4.
Step 4: Solve for xby subtracting 3from both sides: x= 1.
Therefore, the solution to the exponential equation 2x+3 = 16 is x= 1.
24
