# Solving Logarithmic Equations

If you're given an algebraic expression such as $3x+5=14$ , you know to use the inverse of the operations to solve it. First, you would subtract 5 from each side to get $3x=9$ , and then you would divide each side by 3 to isolate the variable. You end up with $x=3$ .

You've probably used addition, subtraction, multiplication, and division to solve algebra problems like that, but you now know that exponents have an inverse operation too: logarithms. If we see an algebra problem with a logarithm, we can rewrite it in exponential form and solve for x (or another variable). In this article, we'll briefly review how to write equations in exponential form and then add logarithms to our algebraic repertoire. Let's get started.

## Solving logarithmic equations: a review of exponential form

As you've previously learned, logarithms are exponents that answer the question, "What power would I need to raise this base number to get the answer I'm looking for?" The small subscript after the log indicates what base we're working with, such that ${\mathrm{log}}_{b}a=x$ is the mathematical way of saying ${b}^{x}=a$ . If there is no subscript, assume a base of 10.

For example, let's say we want to rewrite
${\mathrm{log}}_{6}216=3$
in exponential form. We're raising 6 (the base) to some power since
it's the subscript **b**, 216 is our desired answer **a**, and
3 is the power we're raising 6 to **x**. In exponential form, we
would say 6^3 = 216. The process can get tricky since we don't think
in terms of exponents very often, but it's not too difficult once
you get some practice in. If you'd like a mnemonic to help you out,
remember that b is the base raised to some power x, and a is the
answer we get after doing so. Once you have those 2, the x is the
only number left.

## Let's try solving logarithmic equations

With that refresher on exponential form out of the way, let's try some practice problems. We'll begin with a simple one:

${\mathrm{log}}_{3}x=4$

Converting this equation to exponential form (b^x = a) entails treating the 3 as b, the 4 as x, and the x as a. We'll get:

$3{4}^{=}$

From here, we simply multiply $3\mathrm{\ast}3\mathrm{\ast}3$ to get an answer of 81. We've successfully solved a logarithmic equation.

Next, let's try a slightly harder one:

$\mathrm{ln}x=3$

This one doesn't say "log," but recall that ln is shorthand for logarithms in base e: the irrational constant with multiple applications in calculus. So, we're effectively looking at:

${\mathrm{log}}_{e}x=3$

Our base (b) is **e**, we're raising it to the third power (x),
and the x is the number we get from doing so (a). So:

${e}^{3}=3$

Now, we multiply ${e}^{3}$ . You should punch that into a calculator since e is an irrational number for an answer of $x\approx 20.086$ . We've solved another one.

Let's look at one more example:

$\mathrm{log}\left(x-3\right)=2$

This one doesn't have a subscript after the log, so we assume that b is 10. The 2 is the power we're raising the base to, and x - 3 is our answer a. In exponential form, we're looking at:

${10}^{2}=x-3$

$100=x-3$

Now we just have a linear expression to solve. Adding 3 to both sides isolates the variable, giving us an answer of $x=103$ . The procedure would be similar if we had $3x$ instead of $x-3$ , except that we would use division to isolate the variable.

We need to take our time with problems like these because there is ample opportunity for minor mistakes, but the math involved isn't that bad. As long as you don't try to rush, you can get it.

## Practice problems on logarithmic equations

a. Solve for x : $\mathrm{log}\left(2x+1\right)=3$

The log doesn't have a subscript, so we assume $b=10$ . We're raising that to the third power and setting it equal to $2x+1$ . In exponential form, the equation should look like this:

${10}^{3}=2x+1$

Next, calculate the value of ${10}^{3}$ :

$1000=2x+1$

Subtract 1 from each side:

$999=2x$

Divide each side by 2 to isolate the variable:

$x=499.5$

b. $\mathrm{log}\phantom{\rule{4pt}{0ex}}1000=17+x$

Whenever we see a logarithm with no subscript, we work in base 10 ( $b=10$ ). We can simplify $\mathrm{log}\phantom{\rule{4pt}{0ex}}1000$ to 4 because ${10}^{4}=10000$ , giving us the following equation:

$4=17+x$

We isolate the variable by subtracting 17 from both sides, giving us a final answer of $-13=x$ .

c. Solve for x: ${\mathrm{log}}_{2}\left(3x-2\right)={\mathrm{log}}_{2}\left(2x+6\right)$

Now, this is a fun one. We have logarithms on both sides of the equation, but they have the same base. That means we can cancel out the logarithms on each side of the equation just like any other operation.

Since the bases of the logs are the same, we can ignore them and set the remaining terms equal to each other:

$3x-2=2x+6$

Next, we have to pick one side to isolate the variable on. Let's subtract $2x$ from each side:

$x-2=6$

Add 2 to each side to isolate and solve for x:

$x=8$

d. Solve for x: ${\mathrm{log}}_{9}\phantom{\rule{4pt}{0ex}}(\mathrm{2x}+1)=2$

In this equation, $b=9$ and we're raising it to the second power. We're setting that equal to $2x+1$ , giving us the following equation:

${9}^{2}=2x+1$

$81=2x+1$

Time to start isolating x. First, subtract 1 from each side:

$80=2x$

Next, divide each side by 2:

$40=x$

e. Solve for x: $\mathrm{ln}\phantom{\rule{4pt}{0ex}}\left(3x+8\right)=0$

This question becomes considerably easier once you remember that $\mathrm{ln}\phantom{\rule{4pt}{0ex}}\left(1\right)=0$ . We can replace the 0 with $\mathrm{ln}\phantom{\rule{4pt}{0ex}}\left(1\right)$ to get the following equation:

$\mathrm{ln}\phantom{\rule{4pt}{0ex}}\left(3x+8\right)=\mathrm{ln}\phantom{\rule{4pt}{0ex}}\left(1\right)$

Since the logarithms on each side of the equation have the same base, we can set the internal terms equal to each other:

$3x+8=1$

Subtract 8 from each side:

$3x=-7$

Divide each side by 3 to solve the equation:

$x=-\frac{7}{3}$

## Topics related to the Solving Logarithmic Equations

Logarithmic to Exponential Form

## Flashcards covering the Solving Logarithmic Equations

## Practice tests covering the Solving Logarithmic Equations

College Algebra Diagnostic Tests

## Solving logarithmic equations is easier with Varsity Tutors

While solving logarithmic equations is similar to the algebra students have been performing since they entered high school, the unique properties of logarithms add a few wrinkles to it. If you or your student could use a helping hand with those wrinkles, a 1-on-1 math tutor can provide additional practice problems and explanations until you feel more confident working with logs. The Educational Directors at Varsity Tutors would be happy to provide more info on the benefits of private instruction.

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