$4^{x}+2^{x+1}=18$ Please help me solve?

Question

I tried using logs on both sides or tried treating it as a quadratic but didn't manage to simplify it, Help?:D

Answer 1

Hint:

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

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

So $x=\log_2(-1+\sqrt{19})$.

Answer 2

Hint: $4^x=(2^2)^x=2^{2x}$. This can easily be reduced to a quadratic equation via a neat substitution.

Answer 3

Hint: Let $x = \log_2(y)$, and rewrite your equation as $$(2^x)^2+2(2^x)-18=0$$

Answer 4

$$(2^x)^2+2\cdot (2^x)-18=0$$

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.