Solve for x: $2^x+4^x=8^x$

Question

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?

Answer 1

Guide:

$$y+y^2=y^3$$

$$y^3-y^2-y=0$$

Solve for $y$: $$y(y^2-y-1)=0$$

Answer 2

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 $$\ln(\beta) = \frac{1}{2} \ln(\beta^{2}) = \frac{1}{2} \ln\left(\frac{1}{\alpha^{2}}\right) = - \frac{\ln(\alpha^{2})}{2} = - \ln(\alpha).$$