# Find x: $2^{2x^2}+2^{x^2+2x+2}=2^{4x+5}$

$$2^{2x^2}+2^{x^2+2x+2}=2^{4x+5}$$

I have to find x in this exponential equation.

I tried to write it in another way, like this:

$2^{2x^2}+2\cdot2^{(x+1)^2}=2\cdot2^{4(x+1)}$

But I don't think this would help.

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Divide by $2^{2x^2}$ to get $1 + 2^{-(x-1)^2} \cdot 2^3 = 2^{-2 (x-1)^2} \cdot 2^7$. Substitute $y = 2^{-(x-1)^2}$, and solve the quadratics.
Divide by $2^{4x}$. We get $$2^{2(x^2-2x)}+4\cdot 2^{x^2-2x}=32.$$ Let $w=2^{x^2-2}$. We arrive at the equation $w^2+4w-32=0$, that is, $(w+8)(w-4)=0$. Continue. If we are looking for solutions over the reals, there is almost nothing left to do. Solutions over the complex numbers give rise to some additional complexity.
Since $x^2-2x$ is "almost" a square, we might, during this first process of simplification, complete that square. We might even let $y=x-1$. But we can afford to wait, since we know that after we get to a quadratic in $w$ (any $w$), the rest is mechanical.