SOLVE FOR X:
$\displaystyle
\log_{8}{2^3}+\log_{3}\left(x^2-8\right) = \log_{4}\sqrt{16}+\log_{3}\left(2\right)+2\log_{9} {x}
$
Cool! I didn't know that $\displaystyle \log_{a^2}x=\tfrac{1}{2}\log_a x$.
So $\displaystyle \log_{8}{2^3} = 1$, $\displaystyle \log_{4}\sqrt{16} = 1$, and $\displaystyle 2\log_{9}{x} = 2\log_{3^2}{x} = \frac{2}{2}\log_{3}{x} = \log_{3}{x}$. I get $\displaystyle \log_{3}\left(x^2-8\right) = \log_{3}{x}+\log_{3}{2} \Leftrightarrow \log_{3}\left(x^2-8\right) = \log_{3}{2x}$ which means $\displaystyle 2x = x^2-8$, so we solve the quadratic $\displaystyle x^2-2x-8 = (x-4)(x+2) = 0$ to get $\displaystyle x = 4$ and $\displaystyle x = -2$. Thank you.
Its a consequence of the change of base formula: $\displaystyle \log_{a^2}x=\frac{\ln x}{\ln(a^2)}=\frac{\ln x}{2\ln a}=\frac{1}{2}\left(\frac{\ln x}{\ln a}\right)=\tfrac{1}{2}\log_a x$
But note that $\displaystyle x=-2$ can't be a solution! Do you see why?So $\displaystyle \log_{8}{2^3} = 1$, $\displaystyle \log_{4}\sqrt{16} = 1$, and $\displaystyle 2\log_{9}{x} = 2\log_{3^2}{x} = \frac{2}{2}\log_{3}{x} = \log_{3}{x}$. I get $\displaystyle \log_{3}\left(x^2-8\right) = \log_{3}{x}+\log_{3}{2} \Leftrightarrow \log_{3}\left(x^2-8\right) = \log_{3}{2x}$ which means $\displaystyle 2x = x^2-8$, so we solve the quadratic $\displaystyle x^2-2x-8 = (x-4)(x+2) = 0$ to get $\displaystyle x = 4$ and $\displaystyle x = -2$. Thank you.