Applications of Exponential Functions: Solutions

Here are some solutions to the Exercises to accompany the section Applications of Exponential Functions.  

The Origins of Exponential Functions

  1. Identify the requested quantities in each of the following situations involving exponential functions.
    1. What is the growth factor in the exponential function f(t) = 300(1.25)t.  What is the rate of growth as a percentage?  What is the initial amount?
      Solution
      The growth factor is the base of the exponential, 1.25.  This is 1 + .25, which corresponds to a growth by an additional .25 = 25% each time t increases by 1.  The initial amount is the amount when t = 0, namely, f(0) = 300(1.25)0 = 300·1 = 300.  This confirms the pattern that we saw in the text, that the initial amount equals the constant in front of the exponential.
    2. What is the change factor in the exponential function g(t) = 2(0.85)t.  What is the rate of decrease as a percentage?  What is the initial amount?
      Solution
      The change factor is the base of the exponential, 0.85.  This is 1 - .15, which corresponds to a decrease by a factor of .15 = 15% each time t increases by 1.  As before, the initial amount equals the constant, 2, in front of the exponential.
    3. What is the change factor in the exponential function h(t) = 6(0.5)2t+1.  What is the rate of decrease as a percentage?  What is the initial amount?  Warning: You must use rules of exponents first to put this in the proper form first.
      Solution
      Before describing this function, we must put h(t) in the form h(t) = Cbt.  Using rules of exponents to rewrite the exponent gives h(x) = 6(0.5)2t+1 = 6(0.5)2t(0.5)1 = 3(0.5)2t = 3((0.5)2)t = 3(.25)t.  Now we can see that the change factor is 0.25.  This is 1 - .75, which corresponds to a decrease by a factor of 75% each time t increases by 1.  The initial amount is 3.

    Back to Exercises.

  2. For each of the following situations, write down the corresponding exponential function relating A and t.
    1. You know that the amount, A, of  a radioactive sample decreases by 25% each week, and you begin with 5 grams, give an equation between A and t = the number of weeks after the initial weighing.
      Solution
      Since the initial amount is 5, we can write A = f(t) = 5bt.  Moreover, since it decreases 25% = 0.25 each week, the change factor b = 1 - 0.25 = 0.75.  This implies that the function relating A and t is f(t) = 5(.75)t.
    2. You put $400 in the bank at 8% annual compound interest, give an equation between A = the amount in your account and t = the number of years, assuming you make no withdrawals.
      Solution
      Now the initial amount is 400, so that A = f(t) = 400bt.  Since it increases 8% = 0.08 each year, the change factor b = 1 + 0.08 = 1.08.  This implies that the function relating A and t is f(t) = 400(1.08)t.
       
    3. Babies grow rather quickly from a single cell into a many-celled embryo.  Assuming that A = the number of cells in a baby's body grows by a factor of 8 every minute, give an equation between A and t = the number of minutes after conception.
      Solution
      Since we all start from a single cell, the initial amount is 1, so that A = f(t) = bt.  We are told that he change factor b = 8 (corresponding to a growth rate of 700%/min.).  This may be written in functional terms as A = f(t) = 8t.

    Back to Exercises.

Banking, Interest, and Exponential Functions

  1. Use the correct compound interest formula to determine the following amounts.  Assume that you make no payments or withdrawals.
    1. You put $800 in the bank at 4% compounded weekly, determine the amount you will have in 10 years.
      Solution
      Using the compound interest formula, with I = 800, r = .04, k = 52 (since there are 52 weeks in a year), and t = 10, gives A = 800(1 + .04/52)52·10 » 1193.28.  Note: If you use your calculator, remember to follow the proper order of operations and include all necessary parentheses.
    2. You put $800 in the bank at 4% compounded continuously, determine the amount you will have in 10 years.
      Solution
      Using the continuous compound interest formula, with I = 800, r = .04, and t = 10, gives A = 800e0.04·10 » 1193.46.  Note: This is why, in simple circumstances, most banks generally use continuous compounding, since it gives pretty much the same results with less effort.
    3. You put $200 on a credit card that charges 18% compounded daily, determine the amount you will owe in 3 years.
      Solution
      This time I = 200, r = .18, k = 365 (since there are 365 days in a year), and t = 3, so that you will owe A = 200(1 + .18/365)365·3 » 343.16.  You will pay almost an additional $150 in interest!
    4. You take out a $15000 car loan at 8% compounded monthly, determine the amount you will owe in 5 years.
      Solution
      Now I = 15000, r = .08, k = 12 (since there are 12 months in a year), and t = 5, so that you will owe A = 15000(1 + .08/12)12·5 » 22347.68.  
    5. You want to put some money into a savings account earning 6% compounded daily so that at the end of 4 years, you will have $3000 to take a post-graduation trip to Europe.  Determine how much money you will need to put in the account.  Hint: You will need to leave the initial amount, I, as an unknown in your equation, so that you can simplify and solve for it.
      Solution
      In this case, we have the equation 3000 = I(1 + .06/365)365·4.  Simplifying gives 3000 = 1.27122I, so that I = 3000/1.27122 » 2359.93.

    Back to Exercises.

  2. Use the correct effective interest formula to determine the effective interest rate, reffective, in each situation.  Verify that using the effective interest rate compounded annually gives the same results as in the previous Exercise.
    1. Determine the effective annual rate corresponding to 4% compounded weekly.  Verify that $800 growing annually at this rate gives the same amount as before after 10 years.
      Solution
      Using the first effective interest formula, with r = .04 and k = 52, gives reffective = (1 + .04/52)52 - 1 » 4.0795.  Then, in 10 years, $800 will grow to A = 800(1 + .040795)10 » 1193.28, as we calculated before.
    2. Determine the effective annual rate corresponding to 4% compounded continuously.  Verify that $800 growing annually at this rate gives the same amount as before after 10 years.
      Solution
      Using the second effective interest formula, with r = .04, gives reffective = e0.04 - 1 » 4.081.  Then, in 10 years, $800 will grow to A = 800(1 + .04081)10 » 1193.45, similar to what we calculated before; the discrepancy is due to round-off error.  That is, we rounded the effective interest rate to four digits, which introduced some error into the next calculation.
    3. Determine the effective annual rate corresponding to 18% compounded daily.  Verify that $200 growing annually at this rate gives the same amount as before after 3 years.
      Solution
      This time the effective interest rate turns out to be reffective = (1 + .18/365)365 - 1 » 19.7, so that in 3 years, we can estimate that $200 will grow to A = 200(1 + .197)3 » 343.01.  Here we can see an even larger "round-off error", since we rounded reffective to only three decimal places.
    4. Determine the effective annual rate corresponding to 8% compounded monthly.  Verify that $15000 growing annually at this rate gives the same amount as before after 5 years.
      Solution
      This is the same as the example in the text, so the effective interest rate is (1 + .08/12)12 - 1 » 8.3.  In 5 years, we estimate that a $15000 debt will grow to A = 15000(1 + .083)5 » 22347.70, similar to what we calculated before.

    Back to Exercises.

  3. You might think that the loan examples in the previous Exercise, were not very realistic, since we assumed that we would not make any payments on the loans.  A more realistic formula applies when you pay a fixed amount, p, each period:  

    Notice that this looks very similar to the previous formula, except we take a bit ( ) off the initial loan amount, I, and add it back at the end.  Note: Accountants have a nice, intuitive explanation of this formula in terms of the, so called, "time value of money".

    1. Assume that you put $500 on a credit card that charges 18% compounded monthly, and you make the minimum payment of $10 each month, use this formula to determine how much you would owe in 3 years.
      Solution
      Even after paying $10/mo.(12 mo./yr.)(3 yr.) = $360, you would still owe A = (500 - 10/(.18/12))(1 + .18/12)12·3 + 10/(.18/12) » 381.81!  That is why credit card companies love people that only pay the minimum amount, and why you should try and pay your balance in full each month.
    2. This formula can work to your advantage if you regularly put a fixed amount into an interest-bearing account.  Simply replace p by -p to derive the formula for your balance in a compound interest account if you put in p dollars per period.  
      Solution
    3. Use your formula from part b) to determine how much money you will have in 45 years, if you start with $500 in the bank at 10% compounded weekly and you add $10 per week.
      Solution
      If you start saving at age 20, you would have A = (500 + 10/(.10/52))(1 + .10/52)52·45 + 10/(.10/52) » 516085, more than half a million dollars, when you retire at 65 (i.e., after 45 years), even though you only paid in $10/wk (52 wk./yr.)(45 yr.) = $23,400!  That is why it is a great idea to start a consistent habit of saving for retirement as early as possible.

    Back to Exercises.

Go to Constructing Exponential Models.

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