Exponential with 2 different bases

Question

How to solve for the exponent $x$ in $$8^x + 2^{3x} =\frac{1}{4}?$$ Thank you.

Answer 1

$$8^x+2^{3x}=2^{3x}+2^{3x}=2^{-2}$$ $$2^{3x+1}=2^{-2}$$ $$3x+1=-2$$ $$x=-1$$

Answer 2

Well, we have that:

$$8^x+2^{3x}=\left(2^3\right)^x+\left(2^3\right)^x=2\cdot\left(2^3\right)^x=\frac{1}{4}\space\Longleftrightarrow\space\left(2^3\right)^x=\frac{1}{8}=\left(2^3\right)^{-1}\space\Longleftrightarrow\space x=-1$$