<s>[INST] Solve for x: $2^x+4^x=8^x$ I tried turning $4^x$ and $8^x$ into powers of $2$ and manipulating the equation but could not make progress. What would be the next steps for this  problem? 
 [/INST] Since $4^{x} = 2^{2x}$ and $8^{x} = 2^{3x}$ then the equation becomes
$$2^{x} + 2^{2 x} = 2^{3 x}.$$
Let $t = 2^{x}$ to obtain $t^{3} - t^{2} - t = 0$ or $t (t^{2} - t - 1) = 0$. The quadratic can be factored using $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$ for which
$$t \, (t - \alpha) \, (t - \beta) = 0.$$
This yields that $t = 2^{x}$ has the potential solutions $2^{x} \in \{0, \alpha, \beta\}$. Solving for $x$ in terms of $t$ is $2^{x} = e^{x \ln(2)} = t$,
$$x = \frac{\ln(t)}{\ln(2)}.$$
For the case $t = 0$ then $x = \ln(0)/\ln(2) = - \infty$ and should be eliminated, but is valid since it is essentially saying $0 + 0 = 0$. The other two cases result in 
$$x \in \left\{ \frac{\ln(\alpha)}{\ln(2)}, - \frac{\ln(\alpha)}{\ln(2)} \right\}.$$
The second is obtained by $\alpha \cdot \beta = -1$ and $\alpha^{2} \cdot \beta^{2} =1$ and
$$\ln(\beta) = \frac{1}{2} \ln(\beta^{2}) = \frac{1}{2} \ln\left(\frac{1}{\alpha^{2}}\right) = - \frac{\ln(\alpha^{2})}{2} = - \ln(\alpha).$$
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