Solving a second-degree exponential equation with logarithms

Question

The following equation is given:

$8^{2x} + 8^{x} - 20 = 0$

The objective is to solve for $x$ in terms of the natural logarithm $ln$.

I approach as follows:

$\log_8{(8^{2x})} = \log_8{(-8^{x} + 20)}$

$2x = \log_8{(-8^x + 20)}$

$2x = \dfrac{ \ln{(-8^{x} + 20)} }{ \ln{8} }$

$x = \dfrac{ \ln{(-8^{x} + 20)} }{ 2\ln{8} }$

and at this point I'm unable to proceed.

Answer 1

Hint: Make the substitution $u=8^x$. Now the equation becomes $u^2+u-20=0$.

Answer 2

We have $$(8^x)^2+8^x-20=0$$

$$\implies8^x=\frac{-1\pm\sqrt{1-4(-20)}}2=-5,4$$

For real $x,8^x>0$

Then $8^x=4\implies x\ln8=\ln4$

But $\ln8=\ln(2^3)=3\ln2$ and $\ln4=\ln(2^2)=2\ln2$