Answer
$\left\{7\right\}$
Work Step by Step
Since $\log_b y=x$ implies $y=b^x$, the given equation, $
\log_x\left(\dfrac{1}{49}\right)=-2
,$ implies
\begin{align*}\require{cancel}
x^{-2}&=\dfrac{1}{49}
.\end{align*}
Using the laws of exponents and the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
\dfrac{1}{x^2}&=\dfrac{1}{49}
&(\text{use }a^{-m}=\dfrac{1}{a^m})
\\\\
1(49)&=1(x^2)
&(\text{cross-multiply})
\\
49&=x^2
\\
x^2&=49
.\end{align*}
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{align*}\require{cancel}
x&=\pm\sqrt{49}
\\
x&=\pm7
.\end{align*}
If $x=-7$, the expression $\log_x\left(\dfrac{1}{49}\right)$ becomes $\log_{-7}\left(\dfrac{1}{49}\right)$. This is undefined since in $\log_b x$, $b$ and $x$ are positive numbers.
Hence, the solution set of the equation $
\log_x\left(\dfrac{1}{49}\right)=-2
$, is $
\left\{7\right\}
$.
