$$\small$\text{
Solve for x : $\log_9{(x^2)}=9$
}}$$
$$\small{\text{Formula: $\boxed{~ \log_b{(b^a)}=a~}$}}\\\\
\small{\text{
We have $b=9$, and $a = 9$ so $\log_9{(9^9)}=9$, but $9^9$ is $x^2$
}}\\\\
\small{\text{$
\begin{array}{rcl}
x^2 &=& 9^9\\
x &=& 9^{\frac{9}{2}}\\
x &=& \left(\sqrt{9} \right)^9\\
x &=& \left(\sqrt{3^2} \right)^9\\
x &=& 3^9\\
\mathbf{x} & \mathbf{=} & \mathbf{19683}\\
\\
\hline
\\
\end{array}
$}}\\$$
.
$$\\log_9x^2=9\\\\
2log_9x=9\\\\
log_9x=9/2\\\\
9^{log_9x}=9^{9/2}\\\\
x=9^{9/2}\\\\
x=3^9\\\\
x=19683$$
check:
$${{log}}_{{\mathtt{9}}}{\left({{\mathtt{19\,683}}}^{{\mathtt{2}}}\right)} = {\mathtt{9}}$$
. $$\small$\text{
Solve for x : $\log_9{(x^2)}=9$
}}$$
$$\small{\text{Formula: $\boxed{~ \log_b{(b^a)}=a~}$}}\\\\
\small{\text{
We have $b=9$, and $a = 9$ so $\log_9{(9^9)}=9$, but $9^9$ is $x^2$
}}\\\\
\small{\text{$
\begin{array}{rcl}
x^2 &=& 9^9\\
x &=& 9^{\frac{9}{2}}\\
x &=& \left(\sqrt{9} \right)^9\\
x &=& \left(\sqrt{3^2} \right)^9\\
x &=& 3^9\\
\mathbf{x} & \mathbf{=} & \mathbf{19683}\\
\\
\hline
\\
\end{array}
$}}\\$$