View Full Version : Exponential Equations
shortstuff001
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
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.