Exponential Equations

shortstuff001 · #1 Apr 28, 2009, 03:48 PM

The question is

2^x=2^2x-6

Can you explain how to solve this step by step?

Perito · Apr 28, 2009, 05:48 PM

$2^x=2^{2x}-6$

Later. I started but I don't have time right now to finish -- and I'm a bit confused. I might need to leave this one for galactus.
For what it's worth, X is between 1.58496 and 1.58497, but I worked that out numerically rather than in a closed form.

Zazonker · Apr 29, 2009, 12:29 AM

2^x = 2^2x -6
Rearrange:
2^2x - 2^x - 6 = 0
Factor:
(2^x + 2)(2^x -3) = 0
Set each factor to zero

First factor:
2^x +2 = 0
2^x = - 2 ----- not possible

Second factor:
2^x - 3 = 0
2^x = 3
take the ln of both sides

x ln(2) = ln(3)
x = ln(3)/ln(2)

x= 1.0986/.6931
x=1.58

Perito · Apr 29, 2009, 06:01 AM

Quote:

Originally Posted by Zazonker

$2^x = 2^{2x} -6$

Rearrange:
$2^{2x} - 2^x - 6 = 0$

Factor:
$(2^x + 2)(2^x -3) = 0$

Set each factor to zero

First factor:
$2^x +2 = 0$
$2^x = - 2$ ----- not possible

Second factor:
$2^x - 3 = 0$
$2^x = 3$

take the ln of both sides

$x\, ln(2) = ln(3)$

$x = \frac {ln(3)}{ln(2)}$

$x= \frac {1.0986}{.6931}$

$x\,=\,1.58$

Converted to LaTeX for clarity.