Answer
$x\approx -1.98$
Work Step by Step
RECALL:
$\log_b{x}=y \longrightarrow b^y=x$
Write the equation in exponential form using the rule above to obtain:
$7^{-2}=x+2
\\\frac{1}{7^2}=x+2
\\\frac{1}{49}=x+2$
Solve for $x$ to obtain:
$\begin{array}{ccc}
&\frac{1}{49}-2 &= &x+2-2
\\&\frac{1}{49} -1\frac{49}{49}&=&x
\\&-1\frac{48}{49} &= &x
\\&x &\approx &-1.98\end{array}$