$4^{x}+2^{x+1}=18$ Please help me solve? I tried using logs on both sides or tried treating it as a quadratic but didn't manage to simplify it, Help?:D
Hint:

Let $t=2^x&gt;0$. Then the equation becomes $$t^2+2t-18=0.$$

So $t=-1+\sqrt{19}$. $t\not=-1-\sqrt{19}$ Since $t&gt;0$.

So $x=\log_2(-1+\sqrt{19})$.
$$(2^x)^2+2\cdot (2^x)-18=0$$

Same as $x^2+2x-18=0$ (after substitution).

Use the quadratic formula to find the roots. Set a root equal to $2^x$ then do the log.