# Thread: Powers and Constants

1. ## Powers and Constants

How would I go about solve for x in this equation:

$\displaystyle 2^{(2x+2)}-17\times2^x+4=0$

How would I go about solving it since 17 isn't a power of 2 and I can't figure out a way to take 17?

2. 17 = 16+1 = 2^4 + 1, but you may not even need this.

Let y= 2^x

will that help?

3. $\displaystyle 2^{(2x+2)}-17\times2^x+4=0$

$\displaystyle 2^{2}\times 2^{2x}-17\times2^x+4=0$

$\displaystyle 4\times 2^{2x}-17\times2^x+4=0$

make $\displaystyle a = 2^x$

$\displaystyle 4 a^2-17a+4=0$

Now you have a quadratic. Can you you solve it from here?