# Solving for x in exponent

1. Oct 23, 2015

### cmkluza

1. The problem statement, all variables and given/known data
Solve the equation $4^{x-1} = 2^x + 8$.

2. Relevant equations
Just algebra

3. The attempt at a solution
$4^{x-1} = 2^x + 8$
$2^{2(x-1)} = 2^x + 2^3$
$2^{2x}2^{-2} = 2^x + 2^3$
$\frac{2^{2x}}{2^2} = 2^x + 2^3$
$2^{2x} = 2^22^x + 2^5$
$2^x2^x - 2^22^x = 2^5$
$2^x(2^x - 2^2) = 2^5$

I made a few jumps here and there in my work, but it should make sense. This is about as far as I get before I get stuck. My intuition is to try to get this down to two terms with the same base, 2, so I can just equate their powers, but I can't seem to get rid of any of the terms, so I'm not sure where to go from there. I'm sure I'm missing something obvious, but I can't see it at the moment. Do you have any tips on what a better angle to look at this problem from is? Thanks!

2. Oct 23, 2015

### BvU

Try to think this coming from the other end: what kinds of equations are you supposed to be able to solve ? linear, quadratic, goniometric. So perhaps one of these is carefully hidden in this exercise -- and the quadratic kind is the most likely candidate.
When you then realize that $2^{2x} = (2^x)^2$ and look at your line 3 with that knowledge ...

3. Oct 23, 2015

### Staff: Mentor

$4^{x - 1} = 4^x \cdot 4^{-1} = \frac 1 4 4^x$
Move all terms over to the left side, and multiply both sides by 4.
The equation is quadratic in form, and can be factored, as BvU suggests.

Last edited by a moderator: Oct 24, 2015
4. Oct 23, 2015

### cmkluza

Thanks for your help! I figured it out after realizing it made a quadratic and used the quadratic formula.

5. Oct 24, 2015

### SammyS

Staff Emeritus
Yes. the next to last equation in your OP
$\displaystyle \ 2^x2^x - 2^22^x = 2^5 \$​
can be written as
$\displaystyle \ \left(2^x\right)^2 - 4\cdot2^x = 32 \$ .​