SOLUTION: I have two questions Im really stuck on The first question is: 4^2x = 2^x(x-2) And the second is: 2^x+3 = 3^x-1 If you could he

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I have two questions Im really stuck on The first question is: 4^2x = 2^x(x-2) And the second is: 2^x+3 = 3^x-1 If you could he      Log On


   

Question 66005: I have two questions Im really stuck on
The first question is:
4^2x = 2^x(x-2)
And the second is:
2^x+3 = 3^x-1
If you could help me out with both of these questions that would be greatly appreciated.
Thank you
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
4%5E%282x%29+=+2%5E%28x%28x-2%29%29
%282%5E2%29%5E%282x%29+=+2%5E%28x%28x-2%29%29
2%5E%284x%29+=+2%5E%28x%28x-2%29%29
4x+=+x%28x-2%29
4x+=+x%5E2+-+2x
x%5E2+-+6x+=+0
x%28x+-+6%29+=+0
x+=+0
x+=+6
If you plug x+=+6 in equation
4%5E12+=+2%5E24
correct