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Tutorial on Exponential Functions (2) - Problems

This is a tutorial on how to apply exponential functions to solve problems. Examples with detailled solutions and explanations are included.

Example - Problem 1: The populations of 2 cities grow according to the exponential functions

P1(t) = 100e0.013 t


P2(t) = 110e0.008 t


where P1 and P2 are the populations (in thousands) of cities A and B respectively. t is the time in years such that t is positive and t = 0 corresponds to the year 2004.
When will the populations of the two cities be equal and what will be their populations?

Solution to Problem 1:

  • Let t = t' be the time when P1 and P2 are equal, this leads to the following equation in t'
    100e0.013 t' = 110e0.008 t'

  • Divide both side of the above equation by 100*0.008 t'
    e0.013 t' / e0.008 t' = 110/100

  • Use property of exponential functions ax/ay = ax - y to rewrite the above equation as follows
    e0.013 t'- 0.008 t' = 1.1


  • Simplify the exponent in the left side
    e0.005 t' = 1.1

  • Rewrite the above in logarithmic form (or take the ln of both sides)
    0.005 t' = ln 1.1


  • Solve for t' and round the answer to the nearest unit.
    t' = ln1.1 / 0.005.
    t' is approximately equal to 19 years.
    the year will be 2004 + 19 = 2023.

  • Find the populations when t = t' = 19 years. Use any of the function P1 or P2 since they are equal at t = t'
    P1(t') = 100e0.013*19
    P1(t') is approximately equal to 128 thousands.

    For checking, the graphical solution to the above problem is shown below.

    garphical solution of the above problem.

Matched Problem 1: The populations of 2 cities grow according to the exponential functions

P1(t) = 120e0.011 t


P2(t) = 125e0.007 t


where P1 and P2 are the populations (in thousands) of cities A and B respectively. t is the time in years such that t is positive and t = 0 corresponds to the year 2004.
When will the populations of the two cities be equal and what will be their populations?


Example - Problem 2: The amount A of a radioactive substance decays according to the exponential function

A(t) = A0er t


where A
0 is the initial amount (at t = 0) and t is the time in days (t >= 0). Find r, assuming that the half life of this radioactive substance is 10 days.

Solution to Problem 2:

  • At t = 10 days, the amount A of the substance would be equal to half the initial amount A0 (definition of half life)
    A0er*10 = A0 / 2

  • Divide both side of the above equation by A0
    er*10 = 1 / 2

  • Rewrite the above equation in logarithmic form (or take ln of both sides)
    10r = ln(1/2)


  • Solve for r
    r = 0.1ln(1/2)


  • Approximate r to 3 decimal places.
    r is approximately equal to -0.069.

    For checking, the graph of A(t) = 100e-0.069t is shwown below. Note at t = 0 A = 100 and at t = 10 A is approximately equal to 100/2 = 50.


    garphical solution to problem 2.

Matched Problem 2: The amount A of a radioactive substance decays according to the exponential function

A(t) = A0er t


where A
0 is the initial amount (at t = 0) and t is the time in days (t >= 0). Find r, assuming that the half life of this radioactive substance is 20 days.



More references and links related to the exponential functions in this website.

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