Solve for $x\in R$, $2^x+4^x=8^x$

Using the substitution $2^x=t$, we have

$t+t^2=t^3$

$ \Rightarrow {t^3} - {t^2} - t = 0$

Or, $t({t^2} - t - 1) = 0$

$t \ne 0$ as $t=2^x>0$

So, ${t^2} - t - 1 = 0$

Or, $t = \frac{{1 \pm \sqrt {1 + 4} }}{2} = \frac{{1 \pm \sqrt 5 }}{2}$

Since t > 0, $t = \frac{{1 + \sqrt 5 }}{2}=2^x$

So, $x\log 2 = \log \left( {\frac{{1 + \sqrt 5 }}{2}} \right)$

$ \Rightarrow x = \frac{{\log \left( {\frac{{1 + \sqrt 5 }}{2}} \right)}}{{\log 2}} = \frac{{\log (1 + \sqrt 5 )}}{{\log 2}} - 1$