## How to Solve an Exponential Equation

This post about how to solve an exponential equation is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Solve for x:

$\fn_jvn 5^{3x}=6$

## Solution

You are asked to solve this exponential equation.

How do you get that x out of the exponent so you can solve for it? Consider the relationship between exponents and logarithms:

$\fn_jvn a^b=c<=>\log_a c=b$

where “<=>” means the equations on both sides of the symbol are equivalent; each is true only when the other is true.

Note that the equation on the right-hand side is solved for b, the exponent. Let’s apply that to the equation we’re given. 5 stands in the place of a, 3x in the place of b, and 6 in the place of c. Substituting those values into $\inline \fn_jvn \log_a c=b$ gives:

$\fn_jvn \log_5 6=3x$

Divide both sides by 3 and switch the sides:

$\fn_jvn x=\frac{\log_5 6}{3}$

### Caution

Don’t try to cancel the 6 in the numerator with the 3 in the denominator. $\fn_phv x=\log_{5}6$ is a function carried out on the number 6; it is not a multiple of 6, so it does not cancel with 3.

## Another Way to Solve an Exponential Equation: Use Logs

Take the log of each side.

 $\inline \fn_jvn 5^{3x}=6$ This is the equation that you have to solve for x. $\inline \fn_jvn \log_{5}5^{3x}=\log_{5}6$ If two things are equal, their logs are equal. Why log base 5? Choosing the base that’s in the problem will likely give you the cleanest answer. $\inline \fn_jvn 3x\log _5 5= \log_5 6$ The power rule for logarithms says $\inline \fn_jvn \log a^b=b\log a$. In other words, you can bring that exponent down. Apply this rule to the equation’s left-hand side: $\inline \fn_jvn \log _5 5^{3x}=3x\log _5 5$ $\inline \fn_jvn 3x \cdot 1=\log_5 6$ $\inline \fn_jvn \log _a a= 1$ so $\inline \fn_jvn \log _5 5= 1$. Substitute 1 for $\inline \fn_jvn \log _5 5$. $\fn_jvn&space;{\color{Blue}x=\frac{\log&space;_{5}6}{3}&space;}$ Divide both sides by 3 and there’s the answer.

Related: “Exponential Growth.”

This question is similar to question number 18 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## Question

log base 2 of x

The graph of the log function ${\color{Blue}&space;y=\textup{log}_2x}$ is shown. On the same axes, add the graph of ${\color{Blue}&space;y=\textup{log}_2(2x)}$.

## Solution

How will the graph of ${\color{Blue}\textup{&space;log}_2(2x)}$ compare to the graph of ${\color{Blue}\textup{\textup{}&space;log}_2x}$? Will multiplication by 2 make the graph steeper? Higher?

${\color{Blue}\textup{&space;log}_2(2x)}$ is the logarithm of a product: ${\color{Blue}&space;2x}$. Think of the product rule for logarithms:

${\color{Blue}&space;\textup{log}_a(cd)=\textup{log}_ac+\textup{log}_ad}$

This rule applies whether $c$ and $d$ are variables or constants. Then

${\color{Blue}&space;\textup{log}_2(2x)=\textup{log}_22+\textup{log}_2x}$

${\color{Blue}&space;\textup{log}_22=1}$

so

${\color{Blue}&space;\textup{log}_22x=1+\textup{log}_2x}$

Top curve: log base 2 of x
Bottom curve: log base 2 of 2x

Multiplying the argument (that’s x in this case) of a logarithm by a constant increases the value of the logarithm by a fixed amount. That increase is equal to the logarithm of the constant, the number by which you multiply. The two curves have the same shape. The only change is that one is higher than the other.