Solutions of this equation can be found in which range?

[diredragon]

Homework Statement

$4^{x} - 7*2^{(x-3)/2} = 2^{-x}$
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]

Homework Equations

3. The Attempt at a Solution [/B]
I simplified to
$2^{3x} - 7*2^{(3x - 3)/2} = 1$
$2^{3x} - \frac{7*2^{1/2}}{4}2^{3x/2} - 1 = 0$

 

[haruspex]

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

 

[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 $-1$

 

[haruspex]

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 $-1$

Have you checked both of those satisfy the original equation?

 

[diredragon]

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

 

[haruspex]

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

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

 

[diredragon]

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

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]

 

[haruspex]

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]

Looks right.