How to solve for $x$ in $2^x+4^x=8^x$

Question

So I was bored, and decided to do some math for fun. This was mostly to see if I could still do fairly complex math. After a while, I came up with this to see if I could still do some reasonably simple Algebra involving logarithms:$$\text{Solve for }x\text{: }2^x+4^x=8^x$$which I thought that I might be able to do. Here is my attempt at solving for $x$ in $2^x+4^x=8^x$:

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$$2^x+4^x=8^x$$$$\implies1+2^x=4^x$$$$\because2^x+4^x=8^x\equiv\require{cancel}\cancel{2^x}(1+2^x)=\cancel{2^x}(4^x)\gets1+2^x=4^x$$$$2^x+1-(4^x)=\cancel{4^x-4^x}$$$$-(4^x)+2^x+1=0$$$$2^{2x}-2^x-1=0$$$$(2^x)^2-2^x-1=0$$$$a^2-a-1=0\quad\text{substituting }a\text{ for }2^x$$$$a=\dfrac{1\pm\sqrt{1+4}}{2}$$$$a=\dfrac{1\pm\sqrt5}{2}$$$$2^x=\dfrac{1\pm\sqrt5}{2}$$$$2^x=(1\pm\sqrt5)2^{-1}$$$$x\ln(2)=\ln(1\pm\sqrt5)-\ln(2)$$$$\therefore x=\dfrac{\ln(1\pm\sqrt5)}{\ln(2)}-1$$$$\text{Or }x\approx0.6942419136306,-0.6942419136306+4.3523601i$$

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My question

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Is the solution I have achieved correct, or what could I do to attain the correct solution more easily?

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Mistakes I might have made

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1. Simplifying the logarithms (which honestly is a mistake I always seem to make)
2. The conversion to the quadratic formula (is that even a legal move I made there? I'm not sure. :\)
3. Incorrect tags on my question
4. Incorrectly using math notation

Answer 1

Divide the following by $2^x$ since it is valid for any $x$, $$ 2^x + 4^x = 8^x \rightarrow \frac{2^x}{2^x} + \frac{4^x}{2^x} = \frac{8^x}{2^x} \implies 1 + 2^x =(2^x)^2 $$ Let $y=2^x$, we get $$ 1+y=y^2 \implies y_{1,2} = \frac{1\pm \sqrt{5}}{2} $$ Now we compute $x$; hence, $$ 2^x = \frac{1+ \sqrt{5}}{2} \rightarrow \ln2^x = \ln\left(\frac{1+ \sqrt{5}}{2}\right) \implies x = \dfrac{\ln\left(\frac{1+ \sqrt{5}}{2}\right)}{\ln 2} $$

Because $\sqrt{5} > 1$ and the logarithm is not defined for a negative input, we discard $\frac{1- \sqrt{5}}{2}$; therefore, the equation has one real solution.

Answer 2

$2^x+4^x=8^x$. $u=2^x$. $u+u^2=u^3. \implies u^3-u^2-u=0$

$u(u^2-u-1)=0\implies u\in \{0,\frac{1\pm \sqrt{5}}{2}\}$

So not exactly a real solution, but $x\to -\infty$ works. $x=\ln (\frac{1+\sqrt{5}}{2})/\ln 2 $ is a solution.

$\frac{1-\sqrt{5}}{2}=\frac{\sqrt{5}-1}{2}(e^{\pi i})$

$x=\ln(\frac{1-\sqrt{5}}{2})/\ln (2)=\ln(\frac{\sqrt{5}-1}{2})/\ln(2)+\pi i/\ln(2)$. is another complex solution.