Answer
$x=\dfrac{-97}{49}$
Work Step by Step
RECALL:
$\log_b{x}= y \longrightarrow b^y=x$
Use the rule above to obtain:
$7^{-2}=x+2
\\\dfrac{1}{7^2}=x+2
\\\dfrac{1}{49}=x+2$
Subtract $2$ on both sides of the equation to obtain:
$\dfrac{1}{49}-2=x
\\\dfrac{1}{49} - \dfrac{98}{49}=x
\\\dfrac{-97}{49}=x$
