Find the range of solutions

1. Dec 6, 2015

diredragon

1. The problem statement, all variables and given/known data
$4^{x} - 7*2^{(x-3)/2} = 2^{-x}$
Set of real solutions of this equation is found in which following range:
a) (-9, -2)
b) (0, 3)
c) (-3, 0]
d) (3, 7]
2. Relevant equations
3. The attempt at a solution

I simplified to
$2^{3x} - 7*2^{(3x - 3)/2} = 1$
$2^{3x} - \frac{7*2^{1/2}}{4}2^{3x/2} - 1 = 0$

Last edited by a moderator: Dec 6, 2015
2. Dec 6, 2015

haruspex

There is something that every occurrence of x has in common. This suggests a change of variable.

3. Dec 6, 2015

diredragon

This came to mind
$z^2 = 2^{3x}$
$z^2 - \frac{7*2^{1/2}}{4}z - 1 = 0$
solutions of this equation i named q and t
$q = 2*2^{1/2}$
$t = \frac{-1}{4}2^{1/2}$
I then get two values of $x$, $1$ and $-1$

Last edited: Dec 6, 2015
4. Dec 6, 2015

haruspex

Have you checked both of those satisfy the original equation?

5. Dec 6, 2015

diredragon

I get that neither satisfys the equation. What is the mistake?

6. Dec 6, 2015

haruspex

One does. The other came in because the use of z2 created an ambiguity.

7. Dec 6, 2015

diredragon

Oh I didn't see. $1$ fits. But I don't see how I can get the range which is asked. 1 is found in only one given answer so the solution i guess can only be (0, 3]

Last edited: Dec 6, 2015
8. Dec 6, 2015

Looks right.