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Active Prelude to Calculus

Section 3.5 Properties and applications of logarithmic functions

Logarithms arise as inverses of exponential functions. In addition, we have motivated their development by our desire to solve exponential equations such as \(e^k = 3\) for \(k\text{.}\) Because of the inverse relationship between exponential and logarithmic functions, there are several important properties logarithms have that are analogous to ones held by exponential functions. We will work to develop these properties and then show how they are useful in applied settings.

Preview Activity 3.5.1.

In the following questions, we investigate how \(\log_{10}(a \cdot b)\) can be equivalently written in terms of \(\log_{10}(a)\) and \(\log_{10}(b)\text{.}\)
  1. Write \(10^x \cdot 10^y\) as \(10\) raised to a single power. That is, complete the equation
    \begin{equation*} 10^x \cdot 10^y = 10^{\Box} \end{equation*}
    by filling in the box with an appropriate expression involving \(x\) and \(y\text{.}\)
  2. What is the simplest possible way to write \(\log_{10}10^x\text{?}\) What about the simplest equivalent expression for \(\log_{10}10^y\text{?}\)
  3. Explain why each of the following three equal signs is valid in the sequence of equalities:
    \begin{align*} \log_{10}(10^x \cdot 10^y) &= \log_{10}(10^{x+y})\\ &= x+y\\ &= \log_{10}(10^x) + \log_{10}(10^y)\text{.} \end{align*}
  4. Suppose that \(a\) and \(b\) are positive real numbers, so we can think of \(a\) as \(10^x\) for some real number \(x\) and \(b\) as \(10^y\) for some real number \(y\text{.}\) That is, say that \(a = 10^x\) and \(b = 10^y\text{.}\) What does our work in (c) tell us about \(\log_{10}(ab)\text{?}\)

Subsection 3.5.1 Key properties of logarithms

In Preview Activity 3.5.1, we considered an argument for why \(\log_{10}(ab) = \log_{10}(a) + \log_{10}(b)\) for any choice of positive numbers \(a\) and \(b\text{.}\) In what follows, we develop this and other properties of the natural logarithm function; similar reasoning shows the same properties hold for logarithms of any base.
Let \(a\) and \(b\) be any positive real numbers so that \(x = \ln(a)\) and \(y = \ln(b)\) are both defined. Observe that we can rewrite these two equations using the definition of the natural logarithm so that
\begin{equation*} a = e^x \ \text{ and } \ b = e^y\text{.} \end{equation*}
Using substitution, we can now say that
\begin{equation*} \ln(a \cdot b) = \ln(e^x \cdot e^y)\text{.} \end{equation*}
By exponent rules, we know that \(\ln(e^x \cdot e^y) = \ln(e^{x+y})\text{,}\) and because the natural logarithm and natural exponential function are inverses, \(\ln(e^{x+y}) = x+y\text{.}\) Combining the three most recent equations,
\begin{equation*} \ln(a \cdot b) = x + y\text{.} \end{equation*}
Finally, recalling that \(x = \ln(a)\) and \(y = \ln(b)\text{,}\) we have shown that
\begin{equation*} \ln(a \cdot b) = \ln(a) + \ln(b) \end{equation*}
for any choice of positive real numbers \(a\) and \(b\text{.}\)
A similar property holds for \(\ln(\frac{a}{b})\text{.}\) By nearly the same argument, we can say that
\begin{align*} \ln\left( \frac{a}{b} \right) &= \ln\left( \frac{e^x}{e^y} \right)\\ &= \ln \left( e^{x-y} \right)\\ &= x-y\\ &= \ln(a) - \ln(b)\text{.} \end{align*}
We have thus shown the following general principles.

Logarithms of products and quotients.

For any positive real numbers \(a\) and \(b\text{,}\)
  • \(\displaystyle \ln(a \cdot b) = \ln(a) + \ln(b)\)
  • \(\displaystyle \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b)\)
Because positive integer exponents are a shorthand way to express repeated multiplication, we can use the multiplication rule for logarithms to think about exponents as well. For example,
\begin{equation*} \ln(a^3) = \ln(a \cdot a \cdot a)\text{,} \end{equation*}
and by repeated application of the rule for the natural logarithm of a product, we see
\begin{equation*} \ln(a^3) = \ln(a) + \ln(a) + \ln(a) = 3\ln(a)\text{.} \end{equation*}
A similar argument works to show that for every natural number \(n\text{,}\)
\begin{equation*} \ln(a^n) = n\ln(a)\text{.} \end{equation*}
More sophisticated mathematics can be used to prove that the following property holds for every real number exponent \(t\text{.}\)

Logarithms of exponential expressions.

For any positive real number \(a\) and any real number \(t\text{,}\)
\begin{equation*} \ln(a^t) = t\ln(a)\text{.} \end{equation*}
The rule that \(\ln(a^t) = t\ln(a)\) is extremely powerful: by working with logarithms appropriately, it enables us to move from having a variable in an exponential expression to the variable being part of a linear expression. Moreover, it enables us to solve exponential equations exactly, regardless of the base involved.

Example 3.5.1.

Solve the equation \(7 \cdot 3^t - 1 = 5\) exactly for \(t\text{.}\)
To solve for \(t\text{,}\) we first solve for \(3^t\text{.}\) Adding \(1\) to both sides and dividing by \(7\text{,}\) we find that \(3^t = \frac{6}{7} \text{.}\) Next, we take the natural logarithm of both sides of the equation. Doing so, we have
\begin{equation*} \ln \left( 3^t \right) = \ln \left( \frac{6}{7} \right)\text{.} \end{equation*}
Applying the rule for the logarithm of an exponential expression on the left, we see that \(t \ln(3) = \ln \left( \frac{6}{7} \right) \text{.}\) Both \(\ln(3)\) and \(\ln \left( \frac{6}{7} \right)\) are simply numbers, and thus we conclude that
\begin{equation*} t = \frac{\ln \left( \frac{6}{7} \right)}{\ln(3)}\text{.} \end{equation*}
The approach used in Example 3.5.1 works in a wide range of settings: any time we have an exponential equation of the form \(p \cdot q^t + r = s\text{,}\) we can solve for \(t\) by first isolating the exponential expression \(q^t\) and then by taking the natural logarithm of both sides of the equation.

Activity 3.5.2.

Solve each of the following equations exactly and then find an estimate that is accurate to 5 decimal places.
  1. \(\displaystyle 3^t = 5\)
  2. \(\displaystyle 4 \cdot 2^t - 2 = 3\)
  3. \(\displaystyle 3.7 \cdot (0.9)^{0.3t} + 1.5 = 2.1\)
  4. \(\displaystyle 72 - 30(0.7)^{0.05t} = 60\)
  5. \(\displaystyle \ln(t) = -2\)
  6. \(\displaystyle 3 + 2\log_{10}(t) = 3.5\)

Subsection 3.5.2 The graph of the natural logarithm

As the inverse of the natural exponential function \(E(x) = e^x\text{,}\) we have already established that the natural logarithm \(N(x) = \ln(x)\) has the set of all positive real numbers as its domain and the set of all real numbers as its range. In addition, being the inverse of \(E(x) = e^x\text{,}\) we know that when we plot the natural logarithm and natural exponential functions on the same coordinate axes, their graphs are reflections of one another across the line \(y = x\text{,}\) as seen in Figure 3.5.2 and Figure 3.5.3.
Figure 3.5.2. The natural exponential and natural logarithm functions on the interval \([-3,3]\text{.}\)
Figure 3.5.3. The natural exponential and natural logarithm functions on the interval \([-15,15]\text{.}\)
Indeed, for any point \((a,b)\) that lies on the graph of \(E(x) = e^x\text{,}\) it follows that the point \((b,a)\) lies on the graph of the inverse \(N(x) = \ln(x)\text{.}\) From this, we see several important properties of the graph of the logarithm function.

The graph of \(y = \ln(x)\).

The graph of \(y = \ln(x)\)
  • passes through the point \((1,0)\text{;}\)
  • is always increasing;
  • is always concave down; and
  • increases without bound.
Because the graph of \(E(x) = e^x\) increases more and more rapidly as \(x\) increases, the graph of \(N(x) = \ln(x)\) increases more and more slowly as \(x\) increases. Even though the natural logarithm function grows very slowly, it does grow without bound because we can make \(\ln(x)\) as large as we want by making \(x\) sufficiently large. For instance, if we want \(x\) such that \(\ln(x) = 100\text{,}\) we choose \(x = e^{100}\text{,}\) since \(\ln(e^{100}) = 100\text{.}\)
While the natural exponential function and the natural logarithm (and transformations of these functions) are connected and have certain similar properties, it’s also important to be able to distinguish between behavior that is fundamentally exponential and fundamentally logarithmic.

Activity 3.5.3.

In the questions that follow, we compare and contrast the properties and behaviors of exponential and logarithmic functions.
  1. Let \(f(t) = 1 - e^{-(t-1)}\) and \(g(t) = \ln(t)\text{.}\) Plot each function on the same set of coordinate axes. What properties do the two functions have in common? For what properties do the two functions differ? Consider each function’s domain, range, \(t\)-intercept, \(y\)-intercept, increasing/decreasing behavior, concavity, and long-term behavior.
  2. Let \(h(t) = a - be^{-k(t-c)}\text{,}\) where \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(k\) are positive constants. Describe \(h\) as a transformation of the function \(E(t) = e^t\text{.}\)
  3. Let \(r(t) = a + b\ln(t-c)\text{,}\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are positive constants. Describe \(r\) as a transformation of the function \(L(t) = \ln(t)\text{.}\)
  4. Data for the height of a tree is given in the Table 3.5.4; time \(t\) is measured in years and height is given in feet. At 1 , you can find a Desmos worksheet with this data already input.
    Table 3.5.4. The height of a tree as a function of time \(t\) in years.
    \(t\) 1 2 3 4 5 6 7 8 9 10 11
    \(h(t)\) 6 9.5 13 15 16.5 17.5 18.5 19 19.5 19.7 19.8
    Do you think this data is better modeled by a logarithmic function of form \(p(t) = a + b\ln(t-c)\) or by an exponential function of form \(q(t) = m + ne^{-rt}\text{.}\) Provide reasons based in how the data appears and how you think a tree grows, as well as by experimenting with sliders appropriately in Desmos. (Note: you may need to adjust the upper and lower bounds of several of the sliders in order to match the data well.)

Subsection 3.5.3 Putting logarithms to work

We’ve seen in several different settings that the function \(e^{kt}\) plays a key role in modeling phenomena in the world around us. We also understand that the value of \(k\) controls whether \(e^{kt}\) is increasing (\(k \gt 0\)) or decreasing (\(k \lt 0\)) and how fast the function is increasing or decreasing. As such, we often need to determine the value of \(k\) from data that is presented to us; doing so almost always requires the use of logarithms.

Example 3.5.5.

A population of bacteria cells is growing at a rate proportionate to the number of cells present at a given time \(t\) (in hours). Suppose that the number of cells, \(P\text{,}\) in the population is measured in millions of cells and we know that \(P(0) = 2.475\) and \(P(10) = 4.298\text{.}\) Find a model of the form \(P(t) = Ae^{kt}\) that fits this data and use it to determine the value of \(k\) and how long it will take for the population to reach \(1\) billion cells.
Since the model has form \(P(t) = Ae^{kt}\text{,}\) we know that \(P(0) = A\text{.}\) Because we are given that \(P(0) = 2.475\text{,}\) this shows that \(A = 2.475\text{.}\) To find \(k\text{,}\) we use the fact that \(P(10) = 4.298\text{.}\) Applying this information, \(A = 2.475\text{,}\) and the form of the model, \(P(t) = Ae^{kt}\text{,}\) we see that
\begin{equation*} 4.298 = 2.475 e^{k \cdot 10}\text{.} \end{equation*}
To solve for \(k\text{,}\) we first isolate \(e^{10k}\) by dividing both sides by \(2.475\) to get
\begin{equation*} e^{10k} = \frac{4.298}{2.475} \text{.} \end{equation*}
Taking the natural logarithm of each side, we find
\begin{equation*} 10k = \ln \left( \frac{4.298}{2.475} \right)\text{,} \end{equation*}
and thus \(k = \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right) \approx 0.05519\text{.}\)
To determine how long it takes for the population to reach \(1\) billion cells, we need to solve the equation \(P(t) = 1000\text{.}\) Using our preceding work to find \(A\) and \(k\text{,}\) we know that we need to solve the equation
\begin{equation*} 1000 = 2.475 e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t}\text{.} \end{equation*}
We divide both sides by \(2.475\) to get \(e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t} = \frac{1000}{2.475}\text{,}\) and after taking the natural logarithm of each side, we see
\begin{equation*} \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t = \ln \left( \frac{1000}{2.475} \right)\text{,} \end{equation*}
so that
\begin{equation*} t = \frac{10 \ln \left( \frac{1000}{2.475} \right)}{\ln \left( \frac{4.298}{2.475} \right)} \approx 108.741\text{.} \end{equation*}

Activity 3.5.4.

Solve each of the following equations for the exact value of \(k\text{.}\)
  1. \(\displaystyle 41 = 50e^{-k \cdot 7}\)
  2. \(\displaystyle 65 = 34 + 47e^{-k \cdot 45}\)
  3. \(\displaystyle 7e^{2k-1} + 4 = 32\)
  4. \(\displaystyle \frac{5}{1+2e^{-10k}} = 4\)

Subsection 3.5.4 Summary

  • There are three fundamental rules for exponents given nonzero base \(a\) and exponents \(m\) and \(n\text{:}\)
    \begin{equation*} a^m \cdot a^n = a^{m+n}, \frac{a^m}{a^n} = a^{m-n}, \text{ and } (a^m)^n = a^{mn}\text{.} \end{equation*}
    For logarithms 2 , we have the following analogous structural rules for positive real numbers \(a\) and \(b\) and any real number \(t\text{:}\)
    \begin{equation*} \ln(a \cdot b) = \ln(a) + \ln(b), \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b), \text{ and } \ln(a^t) = t \ln(a)\text{.} \end{equation*}
  • The natural logarithm’s domain is the set of all positive real numbers and its range is the set of all real numbers. Its graph passes through \((1,0)\text{,}\) is always increasing, is always concave down, and increases without bound.
  • Logarithms are very important in determining values that arise in equations of the form
    \begin{equation*} a^b = c\text{,} \end{equation*}
    where \(a\) and \(c\) are known, but \(b\) is not. In this context, we can take the natural logarithm of both sides of the equation to find that
    \begin{equation*} \ln(a^b) = \ln(c) \end{equation*}
    and thus \(b\ln(a) = \ln(c)\text{,}\) so that \(b = \frac{\ln(c)}{\ln(a)}\text{.}\)

Exercises 3.5.5 Exercises


Solve for \(x\text{:}\)
\begin{equation*} 3^{x} = 38 \end{equation*}
\(x =\)


Solve for \(x\text{:}\)
\begin{equation*} 6 \cdot 4^{4 x -4} = 65 \end{equation*}
\(x =\)


Find the solution of the exponential equation
\begin{equation*} 11+5^{5x}=16 \end{equation*}
correct to at least four decimal places.


Find the solution of the exponential equation
\begin{equation*} 1000 (1.04)^{2t} = 50000 \end{equation*}
in terms of logarithms, or correct to four decimal places.


Find the solution of the logarithmic equation
\begin{equation*} 19 - \ln(3-x)=0 \end{equation*}
correct to four decimal places.
Your answer is


For a population that is growing exponentially according to a model of the form \(P(t) = Ae^{kt}\text{,}\) the doubling time is the amount of time that it takes the population to double. For each population described below, assume the function is growing exponentially according to a model \(P(t) = Ae^{kt}\text{,}\) where \(t\) is measured in years.
  1. Suppose that a certain population initially has \(100\) members and doubles after \(3\) years. What are the values of \(A\) and \(k\) in the model?
  2. A different population is observed to satisfy \(P(4) = 250\) and \(P(11) = 500\text{.}\) What is the population’s doubling time? When will \(2000\) members of the population be present?
  3. Another population is observed to have doubling time \(t = 21\text{.}\) What is the value of \(k\) in the model?
  4. How is \(k\) related to a population’s doubling time, regardless of how long the doubling time is?


A new car is purchased for $\(28000\text{.}\) Exactly \(1\) year later, the value of the car is $\(23200\text{.}\) Assume that the car’s value in dollars, \(V\text{,}\) \(t\) years after purchase decays exponentially according to a model of form \(V(t) = Ae^{-kt}\text{.}\)
  1. Determine the exact values of \(A\) and \(k\) in the model.
  2. How many years will it take until the car’s value is $\(10000\text{?}\)
  3. Suppose that rather than having the car’s value decay all the way to $\(0\text{,}\) the lowest dollar amount its value ever approaches is $\(500\text{.}\) Explain why a model of the form \(V(t) = Ae^{-kt} + c\) is more appropriate.
  4. Under the original assumptions (\(V(0) = 28000\) and \(V(1) = 23200\)) along with the condition in (c) that the car’s value will approach $\(500\) in the long-term, determine the exact values of \(A\text{,}\) \(k\text{,}\) and \(c\) in the model \(V(t) = Ae^{-kt} + c\text{.}\) Are the values of \(A\) and \(k\) the same or different from the model explored in (a)? Why?


In Exercise, we explored graphically how the function \(y = \log_b(x)\) can be thought of as a vertical stretch of the nautral logarithm, \(y = \ln(x)\text{.}\) In this exercise, we determine the exact value of the vertical stretch that is needed.
Recall that \(\log_b(x)\) is the power to which we raise \(b\) to get \(x\text{.}\)
  1. Write the equation \(y = \log_b(x)\) as an equivalent equation involving exponents with no logarithms present.
  2. Take the equation you found in (a) and take the natural logarithm of each side.
  3. Use rules and properties of logarithms appropriately to solve the equation from (b) for \(y\text{.}\) Your result here should express \(y\) in terms of \(\ln(x)\) and \(\ln(b)\text{.}\)
  4. Recall that \(y = \log_b(x)\text{.}\) Explain why the following equation (often called the Golden Rule for Logarithms) is true:
    \begin{equation*} \log_b(x) = \frac{\ln(x)}{\ln(b)}\text{.} \end{equation*}
  5. What is the value of \(k\) that allows us to express the function \(y = \log_b(x)\) as a vertical stretch of the function \(y = \ln(x)\text{?}\)