Solving an exponential with three different bases $$2^x+4^x=8^x$$ Solve for $x$.  I reduced the bases to $2$ and try to use logarithms but could not get passed the logarithm of an expression.  When typing into online calculators they say there is no solution but graphing shows an answer.

Set $y  = 2^x$, then you get
\begin{align}
y+y^2= y^3  \ \ \implies \ \ \ y(1+y-y^2) = 0 \ \ \implies \ \ y=0 \ \ \text{ or } \ \ y = \frac{1\pm\sqrt{5}}{2}.
\end{align}
Hence it follows that
\begin{align}
y = 2^x=\frac{1+\sqrt{5}}{2} \ \ \implies \ \ x = \log_2\left(\frac{1+\sqrt{5}}{2}\right).
\end{align}
Note that $y \neq 0$ nor does it equal $(1-\sqrt{5})/2$. 
