Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$

Question

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: $$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {1}{2}} (\frac {1}{2}\log_2 (x^2+7))=-2$$ $$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+1+\log_{1}{2}(\frac {1}{2} \log_2(x^2+7))=-1$$ $$\log_\frac {3}{4} (\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {3}{4}} (\dfrac {3}{4})=-(1+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$ $$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-(\log_{\frac {1}{2}} (\frac {1}{2})+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$ $$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-\log_{\frac {1}{2}} (\frac {1}{4} \log_2 (x^2+7))$$

Answer 1

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 $\ln ( \ln(x^2+7))$.

Answer 2

After your first line you may set $$y= \log_2(x^2+7)$$ for better readability and continue to calculate with $\log_2$.

You get $$\log_{\frac 34}(\frac 13 y) + \log_{\frac 12}\frac 12 y=-2$$

Now, using $\log_b a =\frac{\log_2 a}{\log_2 b}$ you can isolate $\log_2 y$ (I leave the intermediate steps to you):

$$\log_2 y = 2\Leftrightarrow x^2+7 = 16 \Leftrightarrow x=\pm 3$$

Answer 3

I want to do some replacement but it's going to be tricky.

$\log_8 (x^2+7)=a$

$8^a=x^2+7$

$2^{3a}=x^2+7$

$2^{-3a}=(x^2+7)^{-1}$

$(2^{-2})^{\frac{3}{2}a}=(x^2+7)^{-1}$

$\left(\dfrac{1}{4} \right)^{\frac{3}{2}a}=(x^2+7)^{-1}$

$\log_{\frac{1}{4}} (x^2+7)^{-1}=\dfrac{3}{2}a$

The whole thing changes to:

$\log_{\frac{3}{4}} a+\log_{\frac{1}{2}} \dfrac{3}{2}a=-2$

And with the change of base formula (you'll realize the base doesn't matter) then:

$\dfrac{\log a}{\log \dfrac{3}{4}}+\dfrac{\log \dfrac{3}{2}+\log a}{\log \dfrac{1}{2}}=-2$

$\log a \left(\log \dfrac{1}{2}+\log \dfrac{3}{4} \right)+\log \dfrac{3}{4}\log \dfrac{3}{2}=-2 \cdot \log \dfrac{3}{4}\log \dfrac{1}{2}$

$\log a \log \dfrac{3}{8}=\log \dfrac{3}{4}\log \left(\dfrac{1}{2} \right)^{-2}-\log \dfrac{3}{4}\log \dfrac{3}{2}$

$=\log \dfrac{3}{4} \left(\log 4-\log \dfrac{3}{2} \right)=\log \dfrac{3}{4}\log \dfrac{8}{3}$

$\log a=\dfrac{\log \dfrac{3}{4}\log \dfrac{8}{3}}{\log \dfrac{3}{8}}=\dfrac{-1 \cdot \log \dfrac{3}{4}\log \dfrac{3}{8}}{\log \dfrac{3}{8}}$

$=-\log \dfrac{3}{4}=\log \dfrac{4}{3}$

$a=\dfrac{4}{3}$

$8^{\frac{4}{3}}=16=x^2+7$

$x=\pm 3$

I'm tired now. Let me log out.