04 7logapps1

Exponential and logarithmic modelling

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Exponential functions are used to model many physical occurrences such as: growth of cells, population growth, continuously compounded interest, radioactive decay, Newtons's law of cooling.

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Let A be the quantity at time t. … Then: … $A = A_0e^{kt}$,

… … where A0 is a constant and represents the initial value of A. (ie at $t = 0$)

… … The number k is the rate constant of the equation.

… … … Growth: k > 0 … {increasing over time}
… … … Decay: k < 0 … {decreasing over time}

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Example 1

The population of a town was 8,000 at the beginning of 1992 and 15,000 at the end of 1999. Assume the rate of growth is exponential.

Find the rate constant correct to 4 decimal places, and hence

a) … Find the population at the end of 2001.

b) … In what year will the population be double that of 1999?

Solution:

Let
… … P = population of town
… … t = time in years (measured from January 1, 1992)

Exponential population growth can be modelled by: $P = P_0e^{kt}$.

… … At the beginning of 1992,
… … … $t = 0$
… … … $P = 8000$

… … $8000 = P_0e^0$

… … $P_0 = 8000$ … {Showing that P0 is the initial population}

… … So $P = 8000e^{kt}$

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… … To find k, use the other set of information we were given.

… … At the end of 1999,
… … … $t = 8$
… … … $P = 15,000$

… … $15000 = 8000e^{8k}$

… … $e^{8k}=\dfrac{15}{8} \;\; \iff \;\; \log_e \left( \dfrac{15}{8} \right)=8k$

… … $k=\dfrac{1}{8} \log_e \left( \dfrac{15}{8} \right)$

… … $k = 0.07858$

… … The rate of increase of the populaton is 7.9% per annum.

… … hence Exponential growth equation is:

… … $P = 8000e^{0.07858t}$

… … keep longer k value in your calculator for later calculations,
… … but don't write too many decimal places

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a) … Find the population at the end of 2001.

… … … At the end of 2001, t = 10

… … … $P = 8000e^{0.07858×10}$

… … … $P = 17553$

… … … The town's population at the end of 2001 is approximately 17,550.
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b) … In what year will the population be double that of 1999?

… … … When does P = 30,000?

… … … $30000=8000e^{0.07858 \times t}$

… … … $e^{0.07858 \times t}=\frac{30}{8}$

… … … $\log_e \left( \frac{30}{8} \right) =0.07858 \times t$

… … … $t = 16.82$

… … … The population reaches 30,000 in the 17th year, that is during 2008.

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Modelling on the Calculator

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This question could have been answered using the CAS calculator.

… … To find the original equation, use exponential regression:

… … Go to the statistical data entry page and enter the information we are given:

… … … (0, 8000) and
… … … (8, 15000)
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… … … Select Exponential Regression and you will get the equation:

… … … $y=8000e^{0.078576x}$

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… … … Make sure this equation is in the graphing section (y1 = …)

… … … then graph the equation over a suitable domain.
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… … … To find the population after 10 years, go to trace and enter x = 10

… … … To find when the population reaches 30,000:

… … … Draw in a second equation: y2 = 300000

… … … Find the point where the two lines intersect.

… … … Population reaches 30,000 when x = 16.82 (ie during 2008)

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OR

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… … … Obtain $y=8000e^{0.078576x}$ by exponential regression (as above)

a) … solve for y, when x = 10
… … … $P(10) = 17,553$
… … … so population at end of 2001 is 17553
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b) … solve for x, when y = 30000
… … … Pop doubles when t = 16.82
… … … so in 17th year after end of 1991
… … … so population doubles in 2018

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Modelling with Logarithmic Functions

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Logarithmic functions are used to model physical applications such as: magnitude of earthquakes, intensity of sound, acidity of a solution.

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Example 2

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… A solute is being dissolved in water.

… The concentration of the solute in the solution is modelled by $C = \log_e \big(k(t - b) \big)$

… … where C = concentration in moles per litre (M)
… … and t = mixing time (seconds).

… If the concentration is 0.4 M after 5 seconds and 1.2 M after 20 seconds, find:

a) … The values of k and b correct to 3 decimal places

b) … How long does the solution need to be mixed to completely dissolved?

… … … Assume the solute has completely dissoved when the concentration reaches 2 M.
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Solution:

a)$C = \log_e \big( k(t - b) \big)$ … … find k and b

… … … When $t = 5, \; C = 0.4$

… … … $0.4 = \log_e \big(k(5 - b) \big)$ … … (1)
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… … … When $t = 20, \; C = 1.2$

… … … $1.2 = \log_e \big(k(20 - b) \big)$ … … (2)

… … … Use the CAS calculator to solve (1) and (2) simultaneously:

… … … $k = 0.122$,

… … … $b = –7.239$

… … … hence $C = \log_e \big( 0.122(t + 7.239) \big)$
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b) … How long does the solution need to be mixed to completely dissolved?

… … … $C = \log_e \big( 0.122(t + 7.239) \big)$

… … … Need to find t when C = 2

… … … $2 = \log_e \big(0.122(t + 7.239) \big)$

… … … Using the CAS calculator to solve, we get

… … … $t = 53.327$ seconds

… … … The solute will completely dissolve after 54 seconds of mixing.

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