Question

How do you solve `2^(x^2+x) - 4^(1+x) = 0`?

Answer 1

Express both terms as powers of `2` and rearrange to get a quadratic in `x`, giving solutions: `x = -1` or `x = 2`

Explanation:

Add `4^(1+x)` to both sides to get:

`2^(x^2+x) = 4^(1+x) = (2^2)^(1+x) = 2^(2(1+x)) = 2^(2x+2)`

So `x^2+x = 2x+2`

Subtract `2x+2` from both sides to get:

`0 = x^2-x-2 = (x-2)(x+1)`

So `x = -1` or `x = 2`