<s>[INST] Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$ Solve for $x$:  $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$
My Attempt:
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {1}{2}} (\frac {1}{2}\log_2 (x^2+7))=-2$$
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+1+\log_{1}{2}(\frac {1}{2} \log_2(x^2+7))=-1$$
$$\log_\frac {3}{4} (\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {3}{4}} (\dfrac {3}{4})=-(1+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$
$$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-(\log_{\frac {1}{2}} (\frac {1}{2})+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$
$$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-\log_{\frac {1}{2}} (\frac {1}{4} \log_2 (x^2+7))$$
 [/INST] Let's use 
\begin{eqnarray*}
\log_b(a) = \frac{\ln(a)}{\ln(b)}
\end{eqnarray*}
to change everything into natural logarithms.
Your equation becomes
\begin{eqnarray*}
 \frac{\ln \left( \frac{\ln(x^2+7)}{\ln(8)} \right) }{\ln(3/4)} +  \frac{\ln \left( - \frac{\ln(x^2+7)}{\ln(1/4)} \right) }{\ln(1/2)} =-2 \\ 
\end{eqnarray*}
\begin{eqnarray*}
 \frac{\ln ( \ln(x^2+7)) -\ln(\ln(8))  }{\ln(3/4)} +  \frac{  \ln(\ln(x^2+7)) -\ln (\ln(4)) }{\ln(1/2)} =-2 \\ 
\end{eqnarray*}
It is now just linear algebra to isolate $\ln ( \ln(x^2+7))$.
</s>