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## Section4.2Gist of Logarithmic Functions

### Subsection4.2.1A Tool for Solving Exponential Equations

Logarithms are functions that, among other things, can be used to solve exponential equations. If we have an exponential equation in which the variable is in the exponent, we use logarithms to isolate the variable.

By definition, a logarithm is an exponent. In the exponential equation

\begin{equation*} b^{y}=x \end{equation*}

The exponent $y$ is the logarithm so

\begin{equation*} y=\log_{b}(x) \end{equation*}

There are at least three useful rules about logarithms:

The first rule is often used to “bring the exponent down” — after applying the logarithm, the exponent becomes a factor.

The nest two rules are sometimes used to combine two terms involving logarithms, or to split one term into two.

1. $\log_{b}(N^{x})=x \log_{b}(N)$

2. $\log_{b}(AB)=\log_{b}(A)+\log_{b}(B)$

3. $\log_{b}\mathopen{}\left(\frac{A}{B}\right)\mathclose{}=\log_{b}(A)-\log_{b}(B)$

The first rule is perhaps the most useful at this level.

I used to have a pocket knife with $10$ different tools folded into it, but I mostly used only one. Rule 1 is that tool. It allows us to take an exponential expression and “bring down the exponent” so we can isolate it.

If we end up with an equation that looks like

\begin{equation*} (0.337)^{\mathord{?}}=1.272 \end{equation*}

as in, “I wonder what the exponent is?,” we can use a logarithmic function to find the answer.

Although most likely a variable is used instead of a question mark, like

\begin{equation*} (0.337)^{x}=1.272 \end{equation*}

where $x$ is the unknown value we are trying to find.

Using Rule 1 and applying any logarithm to both sides of the equation we can work our way towards isolating the unknown variable $x\text{.}$

\begin{gather*} (0.337)^{x}=1.272\\ \log(0.337)^{x}= \log(1.272)\\ x \log(0.337)= \log(1.272)\\ x=\frac{\log(1.272)}{\log(0.337)}\\ x \approx -0.2212 \end{gather*}

In this case we used $\log\text{,}$ but we get the same answer using the natural logarithm, $\ln\text{,}$ instead. Try it!

It does not matter what base logarithm you use, as long as you use the same $\log_{b}$ on both sides of the equation. However, it does help to use a function that is readily available on most calculators like base $10$ ($\log$), or base $e$ ($\ln$).

Now you try it. Solve an exponential equation using Rule 1.

How many $7 \%$ increases is the same as one $49 \%$ increase?

Rule 1 is great, but if we want to use it as the main tool we have to know when it applies and how to deal with the situation if the rule does not apply.

Solve the equation for $x$

\begin{equation*} 3(2.83)^{x}=15 \end{equation*}

At first glance it seems you can apply Rule 1, but is that right? Rule 1 states

\begin{equation*} \log_{b}(N^{x})=x \log_{b}(N) \end{equation*}

Notice how Rule 1 assumes you have one side written as $N^{x}$ with nothing else in front (no coefficient). The left side of our equation has a coefficient of $3$ in front of the exponential

\begin{equation*} 3(2.83)^{x} \end{equation*}

The constant $3$ is keeping us from using Rule 1. That constant either has to be moved or we have to use another rule (if there is one that applies).

We can fix the situation by rearranging the equation to match the format of Rule 1 or we could use Rule 2 because at least it matches the format of our equation.

Solution

The constant $3$ needs to be moved so only the exponential remains on the left side of the equation.

\begin{equation*} (2.83)^{x}=\frac{15}{3} \end{equation*}

Voilà! Now we have the correct form to use Rule 1. Apply your favorite logarithm to both sides. I'll use $\ln$ this time because I used $\log$ in the last example.

\begin{equation*} \ln (2.83)^{x}= \ln \left( \frac{15}{3} \right) \end{equation*}

The exponent comes down and we have:

\begin{equation*} x \ln(2.83)= \ln \left( \frac{15}{3} \right) \end{equation*}

The last step is to isolate the variable

\begin{equation*} x= \frac{\ln \left( \frac{15}{3} \right)}{\ln(2.83)} \end{equation*}

so that (rounded to three decimals):

\begin{equation*} x \approx 1.547 \end{equation*}

Alternatively, we can use Rule 2. In English Rule 2 says, “The log of a product is equal to the sum of the logs.”

Applying $\ln$ to both sides of our equation, we in fact have the log of the product $3$ times $(2.83)^{x}$ on the left side

\begin{equation*} \ln (3(2.83)^{x})=\ln(15) \end{equation*}

Use Rule 2 to write the left side of the equation as the sum of the logs

\begin{equation*} \ln(3)+\ln\mathopen{}\left(2.83^x\right)\mathclose{}=\ln(15) \end{equation*}

Remember the goal is to isolate the variable $x$ which we can now do since, on the left side, one term has $x$ and the other does not.

Subtracting $\ln(3)$ from both sides we get

\begin{equation*} \ln\mathopen{}\left(2.83^x\right)\mathclose{}=\ln(15)-\ln(3) \end{equation*}

Look, we get to use Rule 1 anyway! On the left we bring down the exponent, $x$

\begin{equation*} x \ln(2.83)=\ln(15)-\ln(3) \end{equation*}

Now isolate $x$

\begin{equation*} x=\frac{\ln(15)-\ln(3)}{\ln(2.83)} \end{equation*}

We get the same answer as before

\begin{equation*} x \approx 1.547 \end{equation*}

Which method did you like better?

Of course we could have just solved the equation $3(2.83)^{x}=15$ graphically and avoided the algebra all together.

###### Remark4.2.5.

A note about solving. In the two algebra examples we were able to find exact solutions:

\begin{equation*} x= \frac{\ln \big( \frac{15}{3} \big)}{\ln(2.83)} \end{equation*}

is exact.

We then entered the exact expressions into a calculator and found approximations rounded to three decimal places. Though graphically may seem quick and easy, most often we can only get an approximation to the answer.

Finally, we could just ask a computer to solve the equation for us. Many websites have tools for solving equations. You can enter an equation like: 3(2.83)^x = 15 and then type solve or click a solve button to see the solutions. You may wish to perform a web search for such a website.

We should all be aware that technology can now do much of what only people could do in the past. And we should also be aware of the current limitations of technology.

In fact, computers can solve all kinds of equations, but so far only people have the ability to set the problems up. So the real question is not necessarily “Can you solve an equation?,”, but “Do you have the right equation?”

### Subsection4.2.2Domain and Range

Because a logarithm is just another kind of function, it has a domain and a range and the graph of a logarithmic function has its own characteristics that make it unique.

We can use our language approach to understand the domain and range of a logarithmic function. Remember the literal translation of the expression $\log(x)$ is “The power of $10$ that gives you $x\text{.}$”

• $\log(100)$ means “The power of $10$ that gives you $100\text{.}$”

Therefore

\begin{equation*} \log(100)=2 \end{equation*}

which says, “The power of $10$ that gives you $100$ is $2$”, since $10^{2}=100\text{.}$

• $\log_{3}(81)$ means, “The power of $3$ that gives you $81\text{.}$”

Therefore

\begin{equation*} \log_{3}(81)=4 \end{equation*}

which says, “The power of $3$ that gives you $81$ is $4$”, since $3^{4}=81$

Recall the domain of a function $f$ is the set of all possible inputs the function can use (values that make sense for the function).

Let

\begin{equation*} f(x)=\log_{3}(x) \end{equation*}

where $x$ is the input.

Consider negative input values, like $x=-9$ and think about the meaning of $\log_{3}(-9)\text{:}$

“What power of $3$ gives you $-9\text{?}$”

The statement should make no sense. You cannot get a negative result if all you have is positive $3$'s multiplied together. At least one has to be negative!

Since any negative input leads us to the same conclusion, it turns out logarithms cannot accept negative inputs.

For a similar reason logarithmic functions cannot have zero as an input. If you ask “How many factors of $3$ does it take to get $0\text{?}$” or just $3^{\mathord{?}}=0\text{,}$ it should be clear there is no answer.

Therefore, the domain of any logarithmic function is all inputs that are greater than or equal to zero.

Domain of $\log_{b}(x)\text{:}$ $x \gt 0\text{.}$

For the range we must consider all the possible outputs of the function. In the case of the logarithm, an output is the exponent value needed to get a particular input

\begin{equation*} \log_{b}(\text{input})=\text{output} \end{equation*}

Consider again $y=\log_{3}(x)\text{.}$

For any input $x$ how many $3$'s does it take to get that $x$-value? The answer (output) is the $y$-value.

 Input Output (exponent) Reason $\frac{1}{27}$ $-3$ Because $3^{-3}=\frac{1}{27}$ $\frac{1}{9}$ $-2$ Because $3^{-2}=\frac{1}{9}$ $\frac{1}{3}$ $-1$ Because $3^{-1}=\frac{1}{3}$ $1$ $0$ Because $3^{0}=1$ $3$ $1$ Because $3^{1}=3$ $9$ $2$ Because $3^{2}=9$

For inputs between zero and one, $0 \lt x \lt 1\text{,}$ the results are negative, so $y \lt 0\text{.}$

For inputs greater than one, $x \gt 1\text{,}$ the results are always positive, so $y \gt 0$

If the input is $x=1$ the result is always zero

\begin{equation*} \log_{3}(1)=0 \end{equation*}

because, “The power of $3$ that gives you $1$” is the $0$ power.

After a few tries we should begin to realize the range of $\log_{3}(x)$ is all real numbers.

It turns out the range for any base of a log function like $\log_{b}(x)$ is all real numbers.

Range of $\log_{b}(x)\text{:}$ All real numbers.

### Subsection4.2.3The Graph of a Logarithmic Function

Just like exponential functions have horizontal asymptotes as in Figure 4.2.8, logarithm functions have vertical asymptotes as in Figure 4.2.9.

The vertical asymptote is like a vertical boundary that the function cannot cross. In the case of the logarithm $y=\log_{b}(x)$ the asymptote is the $y$-axis with equation $x=0$ and the function only exists to the right of this boundary.

This supports our assertion that the domain of $y=\log_{b}(x)$ is all $x \gt 0$ (to the right of the $y$-axis).

### Subsection4.2.4Change of Base

So far, when evaluating expressions like $\log_{3}(9)$ it seems we need to know the answer before we evaluate. How do you evaluate an expression when you don't already know the answer?

Evaluate $\log_{3}(0.1822)\text{.}$

We know the answer should be negative because we are trying to get $0.1822$ which is less than $1\text{.}$

Your calculator may already have a “log of any base” option. Today, in the 21st century we can also find the answer, for free, on the web. For instance, go to Desmos.com. Type log_(0.1822) into one of the cells. You should see the answer as $\log_{3}(0.1822) \approx -1.5498$ and you can get more decimal places if you want.

But before the web, it used to be that calculators only had one or two options for logarithms, (and before calculators there were things called “slide rules” and books with pages and pages of tables of logarithms).

Here is an “old school” method of evaluating a logarithm expression. The method is called “change of base” which means you change the base from whatever it is to either base $10\text{,}$ $\log\text{,}$ or base $e\text{,}$ $\ln\text{,}$ because most calculators have those built in.

Set up an equation in which $y$ is the answer you want. Then use the definition of the logarithm to turn the $log_{b}$ equation into an exponential equation. The last step is to solve the exponential equation using the most convenient logarithm at your disposal (most likely $log$ or $ln$). In this example we use log-base-10, $\log\text{.}$

\begin{align*} \log_{3}(0.1822)\amp=y\\ 0.1822\amp=3^{y}\\ \log(0.1822)\amp=\log(3^{y})\\ \log(0.1822)\amp=y\log(3)\\ y\amp=\frac{\log(0.1822)}{\log(3)}\\ y\amp\approx -1.5498 \end{align*}

### Subsection4.2.5Setting Up Problems (Only to Tear Them Down)

We conclude this section by working through some examples.

Solve an exponential equation

Solve an exponential equation

Solve an exponential equation

Solve an exponential equation

Solve a logarithmic equation

### Subsection4.2.6Conclusion

Student Learning Outcome: Determine the domain and range of functions.

Student Learning Outcome: Sketch the graph of a logarithmic function.

Student Learning Outcome: Identify a vertical/horizontal asymptote from a graph, numerical model, verbal description or equation.

Student Learning Outcome: Solve equations algebraically using properties of exponents and logarithms.

Student Learning Outcome: Use the characteristics of basic functions, especially slope, intercepts, rate of change, percent change and average change to answer questions in application situations, to write equations and to create graphs by hand and on the calculator.

### Exercises4.2.7Exercises

Definition of the Logarithm

###### 4.

Solving Exponential Equations with Logarithms

###### 14.

Solving Logarithmic Equations

###### 22.

Additional Problems

Applications