Math 1101

20. Your savings account grows by continuous compounding of interest. The yearly percentage

growth rate for the account balance M (in dollars) is 4.5%, and your initial

investment is $250.

(a) Find a formula for M as a function of the time t in years since you opened the

account.

(b) What is the monthly percentage growth rate for this account?

21. You deposit some money in a savings account and let the deposited money collect

interest. The following table shows your balance B (in dollars) as a function of time

t, measured in years since your initial deposit.

t 0 1 2 3

B 125.00 130.75 136.76 143.06

(a) By calculating ratios, show that the data is exponential. (Round the ratios to 3

decimal places.)

(b) Find a formula for the exponential function giving the balance B as a function of

time t.

(c) What is the yearly percentage growth rate for this account?

(d) What is the decade percentage growth rate for this account?

(e) Use a table of values on the calculator to estimate how long it will take for your

initial investment to double in size.

Sample Test Questions for Chapter 4 11

22. You are saving money with the hope of buying a new car after several years. Both the

balance in your savings account and the cost of the car grow exponentially over time.

Let t be time in years since the start of 1998.

(a) The cost C (in dollars) of the car at the start of 1998 is $14,000, and the yearly

growth factor for the cost is 1.02. Find a formula for C as a function of t.

(b) At the start of 1998 you invest $12,000 in a savings account. The yearly percentage

growth rate for the account balance B (in dollars) is 4.3%.

i. Find the yearly growth factor for B.

ii. Find a formula for B as a function of t.

iii. Use your answer to Part (a) and to Part (ii) above to determine at what time

your account balance will be large enough so that you are able to afford the

new car.

23. An earthquake in California measured 3.4 on the Richter scale.

(a) An earthquake in Oklahoma measured 1.2 on the Richter scale. How did the

power of the Oklahoma quake compare with that of the California quake?

(b) An earthquake in Chile was 3 times as powerful as the California quake. What

Richter scale reading did the Chilean earthquake have?

24. Recall that pH is a measure of the acidity of a solution and that lower pH values indicate

a more acidic solution. If H is the concentration of hydrogen ions in the solution

(measured in moles per meter of solution), then

pH = −logH.

(a) Solution 1 has a pH of 3. Solution 2 is twice as acidic as solution 1 (meaning that

the concentration of hydrogen ions in solution 2 is twice that of solution 1). What

is the pH of solution 2?

(b) Solution 3 has a pH of 7. How do the concentrations of hydrogen ions for solutions

1 and 3 compare?

12 Sample Test Questions for Chapter 4

25. Recall that

decibels = 10 log(relative intensity).

(a) Sound 1 has a decibel level of 40. Sound 2 has twice the relative intensity of sound

1. What decibel level does sound 2 have?

(b) Sound 3 has a decibel level of 60. How do the relative intensities of sound 1 and

sound 3 compare?

26. Suppose logK = 4.1.

(a) What is the value of log(3K)?

(b) If log L = 7, how do K and L compare?

(c) Find the value of K.

Answers

20. (a) M is an exponential function of the time t and so M = Pat where P is the initial

value, 250, and a = 1 + r = 1.045, so the formula is M = 250 × 1.045t.

(b) To calculate the monthly percentage growth rate we use the growth factors. The

monthly growth factor is 1.0451/12 = 1.0037, so the monthly percentage growth

rate is 0.37%.

21. (a) Since the ratios are all equal to 1.046 (to three decimal places), the data are exponential.

(b) The exponential function giving the balance B as a function of time t is B = Pat

where P is the initial value, 125, and a is the ratio, 1.046, so the formula is B =

125 × 1.046t.

(c) Since the growth factor is 1.046, the yearly percentage growth rate for this account

is 4.6%.

(d) To calculate the decade percentage growth rate we use the growth factors. The

decade growth factor is 1.04610 = 1.5679, so the decade percentage growth rate is

56.79%.

(e) Entering the formula from Part (b) and scrolling for 125 × 2 = 250, the initial

investment doubles in between 15 and 16 years (more precisely: in 15.41 years).

8 Sample Test Answers for Chapter 4

22. (a) Since the initial value of cost C is 14,000, and the yearly growth factor is 1.02, a

formula for C as a function of t is C = 14,000 × 1.02t.

(b) i. Since the yearly percentage growth rate is 4.3%, the yearly growth factor for

B is a = 1 + r = 1 + 0.043 = 1.043.

ii. Since the initial value of B is 12,000, a formula for B is B = 12,000 × 1.043t.

iii. The account balance will be large enough to afford the new car when B = C,

that is when 12,000 × 1.043t = 14,000 × 1.02t. Solving, we find that t = 6.91

years, or near the end of 2004.

23. (a) The Oklahoma earthquake measured 3.4 − 1.2 = 2.2 lower on the Richter scale

than the California earthquake. The Oklahoma quake is therefore 10−2.2 = 0.0063

times as powerful as the California quake.

(b) If the earthquake in Chile was 3 times as powerful as the California quake, then

it registers t points higher on the Richter scale where 10t = 3. Solving, we find

that t = 0.48 or about 0.5, so the Richer scale reading of the Chilean earthquake is

3.4 + 0.5 = 3.9.

24. (a) Since the concentration of solution 2 is twice that of solution 1, the logarithm of

the concentration of solution 2 is t more than that of solution 1 where 10t = 2, and

so the pH of solution 2 is t less than that of solution 1 (because of the minus sign).

Solving for t, we find t = 0.3 and so the pH of solution 2 is 3 − 0.3 = 2.7.

(b) Since solution 3 has a pH of 7, which is 4 more than the pH of solution 1, then the

logarithm of the concentration of solution 3 is 4 less than that solution 1 (because

of the minus sign). Thus the concentration of hydrogren ions for solution 3 is

10−4 = 0.0001 times that of solution 1.

25. (a) Since sound 2 has twice the relative intensity of sound 1, the logarithm of that the

relative intensity of sound 2 is t more that of sound 1, where 10t = 2, so t = 0.3.

Thus the decibel level of sound 2 is 10t = 10 × 0.3 = 3 more than that sound 1, so

that decibel level of sound 2 is 40 + 3 = 43.

(b) Since sound 3 has a decibel level of 60, its decibel level is 20 more than that of

sound 1 and so the logarithm of the relative intensity of sound is 20/10 = 2 times

that of sound 1. Thus the relative intensity of sound 3 is 102 = 100 times that of

sound 1.

26. (a) The value of log(3K) is t more than logK, where 10t = 3. Since t = 0.48, or about

0.5, log(3K) = logK + 0.5 = 4.1 + 0.5 = 4.6.

Sample Test Answers for Chapter 4 9

(b) If log L = 7, then log L is 7 − 4.1 = 2.9 larger than logK, so L is 102.9 = 794.33

times K.

(c) Since logK = 4.1, K = 104.1 = 12,589.25.