Answer
$x=\pm3$
Work Step by Step
Since $\log_b y=x$ implies $y=b^x$, the given equation, $
\log_2(x^2+7)=4
$, implies
\begin{align*}\require{cancel}
x^2+7&=2^4
\\
x^2+7&=16
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
x^2&=16-7
\\
x^2&=9
.\end{align*}
Taking the square root of both sides, the equation above is equivalent to
\begin{align*}\require{cancel}
x&=\pm\sqrt{9}
\\
x&=\pm3
.\end{align*}
Hence, the solution to the equation $
\log_2(x^2+7)=4
$ is $
x=\pm3
$.
