# How do you solve log_4 (3x+1) = 2?

##### 1 Answer
Dec 20, 2015

x = 5. I solved it by sight ... but how?

#### Explanation:

To solve any equation for x, you want to do the same thing to both sides until you isolate the x on the left side.

First undo the log expression by raising 4 to both sides:

${\log}_{4} \left(3 x + 1\right) = 2$

${4}^{{\log}_{4} \left(3 x + 1\right)} = {4}^{2}$ and by the inverse property we get

$3 x + 1 = 16$ which easily solves to $x = 5.$

To check your answer put in $x = 5$ and get

${\log}_{4} \left(3 \left(5\right) + 1\right) = {\log}_{4} \left(16\right) = 2$ because ${4}^{2} = 16.$

/ dansmath strikes again! \