Solve the equation $e\log_4 \left(\sqrt 7^{2^x}\right) = 3^{4x}$

Question

Can anyone help me to solve this equation: $$e\log_4 \left(\sqrt 7^{2^x}\right) = 3^{4x}$$ I tried to solve it graphically using GeoGebra to get the intersection point with the $x$-axis $(y=0)$ and I got $x = 0.13885$ but I cannot prove this algebraically.

Answer 1

Just rewrite it $$2^xe\log_4 \sqrt{7} = 81^x \Leftrightarrow e\log_4 \sqrt{7} = \left(\frac{81}{2}\right)^x$$

Now, take, for example, the natural logarithm:

$$x = \frac{\ln e+\ln \left(\frac{\ln\sqrt{7}}{\ln 4} \right)}{\ln 81 - \ln 2}= \frac{1+\ln\ln\sqrt{7}-\ln \ln 4}{\ln 81 - \ln 2}$$

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