Section 4.3: Exponential Models and Logarithms

In the previous section, we derived a number of algebraic properties of logarithms and saw how they allow us to solve exponential equations.  However, the equations that we solved were presented out of context.  In this section, we will see how such equations arise naturally, when we re-examine the exponential models which we studied earlier and investigate them more deeply.  We will also explain the correspondence between exponential models and linear models, and discuss how researchers determine such models, in practice.

Exponential Models and Exponential Equations

We have seen that many different types of quantities, such as bank balances, populations, drugs, radioactive materials, temperatures, etc., grow or decrease over time according to exponential models.  We have learned how to determine the exact mathematical expressions (involving exponentials) which model these quantities over time.  For example, we know that, if $200 is placed in a savings account at 8% interest compounded monthly, then the amount of money in the account, A, after t years is given by the function, A = f(t) = 200(1 + .08/12)^12t.  

When we first looked at such models, we focused our attention on the output, A, of such exponential functions for a given value of the input, t.  Now that we know about logarithms, we can now ask questions about the input.  For example, we may ask when our account will reach $500; that is, if A = 500, we would like to solve for t.  This means that we must solve the equation 500 = f(t) = 200(1 + .08/12)^12t; in functional terms, we must evaluate the inverse, t = f ^-1(500).  As an inverse of an exponential function, we have seen how to solve such an equation using logarithms, in the usual way, by working "outside-in" applying inverse operations:

• divide by 200, to obtain 500/200 = 200(1 + .08/12)^12t/200 = (1 + .08/12)^12t,
• apply the logarithm function with base b = 1 + .08/12, to obtain log_{(1 + .08/12)}(500/200) = log_{(1 + .08/12)}(1 + .08/12)^12t) = 12t, and
• divide by 12, to obtain (log_{(1 + .08/12)}(500/200))/12 = t = f ^-1(500).

To evaluate this expression, we would then use the change of base formula to write this as (log(500/200)/log(1 + .08/12))/12 » 11.5.  That is, it will take 11.5 years for our account to reach $500.

Since it is common to make mistakes either in algebra (when solving the equation) or arithmetic (when entering a complex expression into our calculator), it is a good habit to use a graph to estimate the answer, as well.  For example, we can use XFunctions to see this result graphically.  Select f and verify that (11.5, 500) is approximately a point on the graph:

Either XFunctions or our calculator shows that f(11.5) = 200(1 + .08/12)^12·11.5  » 500, to three decimal places.

Notice how arriving at an answer to this question required three rather different types of thought processes:

• First, we needed to determine the appropriate mathematical model A = f(t) = I(1 + r/k)^kt, corresponding to our question.
• Then, we identified and substituted the values of the variables which were given in the statement of the question, A = 500, I = 200, r = 0.08, and k = 12, to obtain an equation in a single unknown, 500 = 200(1 + .08/12)^12t. 
• Finally, we used algebra to solve for, t, applying logarithmic functions at appropriate points.

We can apply this same strategy to determine the rate of continuous interest we would need to earn so that an initial deposit of $2000 would grow to $3500 in four years:

• This uses the continuous interest model, A = g(t) = I e^rt.
• The statement of the question gives, A = 3500, I = 2000, and t = 4, yielding the equation, 3500 = 2000 e^4r.  Notice that this time the unknown is the interest rate, r.
• Dividing by 2000, applying ln, and dividing by 4 gives, r = ln(3500/2000)/4  » 0.14, that is, we would need to earn about 14% interest.

These examples both yielded equations similar to the first example in the previous section.  We can easily imagine situations that lead to the second type of equation that we learned to solve.  For example, in the 1990's the U.S. Census showed that, while California had a higher population (about 30 million people) than Texas (about 17 million), Texas' population was growing at a faster rate (about 2% per year) than California's (about 1.3%).  We might ask, if these population trends continue, when will the population of Texas catch up to that of California.  Proceeding as before:

• We recognize that this question again involves the continuous growth model, A = I e^rt, except we have two different functions, , and , for the population of California and Texas, respectively.
• The statement of the question gives, I_C = 30, r_C = 0.013, I_T = 17, and r_T = 0.02.  Although we do not know A_C or A_T, since we are only interested in when these populations are equal, we want to solve the equation A_C = A_T, that is, 30e^0.013t = 17e^0.02t, for t.
• We then proceed as in the second example in the previous section:

  Equation	Reason
  ln(30e0.013t) = ln(17e0.02t)	Apply ln to both sides.
  ln(30) + ln(e0.013t) = ln(17) + ln(e0.02t)	Use the product rule for logarithms.
  ln(30) + 0.013t = ln(17) + 0.02t	Use the inverse rule.
  ln(30) + 0.013t - 0.013t - ln(17) = ln(17) + 0.02t - 0.013t - ln(17)	Subtract terms from both sides.
  ln(30) - ln(17) = 0.007t	Simplify
  (ln(30) - ln(17))/0.7 = t	Divide both sides by the coefficient of x.

  so that t » 81.  That is, if current population trends continue, we can expect the population of Texas to equal and surpass that of California in around the year 2071.

Practice setting up and solving exponential equations by completing the following Exercises.  

Exponential vs. Linear Models

Almost all of the exponential models we have considered to this point (except for Newton's Law of Cooling) have been of the form y = f(x) = Cb^x, and we have seen how to interpret the values for C and b.  Their interpretation was very similar to that of the vertical intercept and slope, respectively, of a linear function.  Moreover, the formula for the change factor, b, was very reminiscent of the slope formula for linear functions.  This is no accident, and the connection between the two types of models can be seen very clearly using logarithms.

If we consider the composite of a logarithmic and exponential function:

we can see that the result is a linear function, with slope, log(b), and vertical intercept, log(C).  Researchers often use this observation to determine the change factor, b, in an exponential model.  For example, if we know that f goes through the two points:

and compute the slope of the resulting line, (log(y₁) - log(y₂))/(x₁ - x₂) = log(y₁/y₂)/(x₁ - x₂), by the quotient rule  Since we know that this equals log(b), we may solve for the change factor, b, by applying the exponential base 10 to both sides:

we obtain the same formula for the change factor as before.

While this is reassuring, this is nothing new.  However, this technique is most useful to researchers when they have many more than simply two data points.  For example, the population of South Central Asia (which includes Afghanistan, Bangladesh, Bhutan, India, Kazakhstan, Kyrgyzstan, Maldives, Nepal, Pakistan, Sri Lanka, Tajikistan, Turkmenistan, and Uzbekistan) has been growing steadily over the last half of the 20th century:

If we plot the log of the population against the year:

we can that the result is approximately a straight line, so that we can conclude that the population is growing exponentially.  Note: This is called a "semi-log" plot; it is common in experimental sciences to use special graph paper (with logarithmically spaced vertical grid lines) to achieve the this effect.  Aside: This property of exponential functions is why so many exponentially growing quantities, such as loudness of sound, ionic concentration in acids, and the destructive force of an earthquake are usually converted logarithmic units (decibels, ph level, and the Richter scale, respectively).

While these points are not exactly in a straight line, we may use a technique from statistics called "linear regression" to estimate the slope of the line that best fits all the data points.  Even if we do not know statistics, we can obtain a rough estimate this line of best fit simply using a ruler:

Estimating the slope from the first and fifth data points yields a value of approximately (8.7 - 9.07)/(1950 - 1990) » 0.00925 for the slope.  This implies that the growth factor for this population is approximately 10^0.00925 » 1.0215; that is, the population of South Central Asia has been growing steadily at a rate of about 2.15% per year over the last 50 years.  Note: While we cannot read the value 9.07 directly off the graph, we know that the value at this point is log(1178234564) » 9.07; using this value our rough estimate actually agrees with that obtained by statistical analysis (2.148%) to three decimal places.

Notice that a slight modification of this technique works for power functions of the form, y = g(x) = Cx^p, since  .  In this case, if we take log of both the inputs and outputs, we obtain a linear function with slope equal to the exponent, p.  For example, in macro-economics, it is often assumed that the total level of economic production, P, is a power function, P = g(K) = CK^p, of the amount of capital investment, K (where the constant, C, depends on other inputs to production, such as labor).  It is then a problem in Econometrics (namely, statistics for economists) to determine the exponent, p, based on the production data of the economy.

For example, if we have the following data:

we would plot the log of capital investment against the log of the measure of total production:

This is not nearly as "straight" as the previous example, but does approximate a straight line:

Note: This is naturally called a "log-log" plot.  Since this line was simply drawn in by hand, it is not particularly accurate, so we cannot be too precise about estimating its slope.  However, the ends of the line look to be at around (4.2, 4.25) and (4.6, 4.55), so the slope is approximately (4.25 - 4.55)/(4.2 - 4.6) = 3/4.  This means that we would estimate the level of production to be proportional to the level of capital investment raised to the three-quarters power, P = g(K) = CK^3/4.  Note: The statistical process of "linear regression" gives reasonable values for both the slope and vertical intercept, so that we could determine the constant of proportionality, C, as well.

To summarize:

Practice estimating exponential and power function models by completing the following Exercises.  

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