## 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 with Logs

Take the log of each side.

 $\inline \fn_jvn 5^{3x}=6$$\inline \fn_jvn 5^{3x}=6$ This is the equation that you have to solve for x. $\inline \fn_jvn \log_{5}5^{3x}=\log_{5}6$$\inline \fn_jvn \log_{5}5^{3x}=\log_{5}6$ If two things are equal, their logs are equal. Why log base 5? Any base is valid; 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$$\inline \fn_jvn 3x\log _5 5= \log_5 6$ The power rule for logarithms says $\inline \fn_jvn \log a^b=b\log a$$\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$$\inline \fn_jvn \log _5 5^{3x}=3x\log _5 5$ $\inline \fn_jvn 3x \cdot 1=\log_5 6$$\inline \fn_jvn 3x \cdot 1=\log_5 6$ $\inline \fn_jvn \log _a a= 1$$\inline \fn_jvn \log _a a= 1$ so $\inline \fn_jvn \log _5 5= 1$$\inline \fn_jvn \log _5 5= 1$. Substitute 1 for $\inline \fn_jvn \log _5 5$$\inline \fn_jvn \log _5 5$. $\inline \fn_jvn x=\log_5 6/3$$\inline \fn_jvn x=\log_5 6/3$ 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.