SOLUTION: Please help me solve this question - I've been trying to set the bases equal. I know that the answer is x=3, but I can't quite get there with showing my work. 2*(2^2x)=(4^x)+64

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help me solve this question - I've been trying to set the bases equal. I know that the answer is x=3, but I can't quite get there with showing my work. 2*(2^2x)=(4^x)+64       Log On


   

Question 1074918: Please help me solve this question - I've been trying to set the bases equal. I know that the answer is x=3, but I can't quite get there with showing my work.
2*(2^2x)=(4^x)+64
Thank you
Found 3 solutions by ikleyn, Fombitz, MathTherapy:
Answer by ikleyn(52847) About Me  (Show Source):
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
4%5Ex=2%5E%282x%29
So,
2%2A2%5E%282x%29=2%5E%282x%29%2B64
Take it from there.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Please help me solve this question - I've been trying to set the bases equal. I know that the answer is x=3, but I can't quite get there with showing my work.
2*(2^2x)=(4^x)+64
Thank you
2%282%5E%282x%29%29+=+4%5Ex+%2B+64

2%282%5E%282x%29%29+-+4%5Ex+=+64 ------ Subtracting 4%5Ex from each side

2%282%5E%282x%29%29+-+%282%5E2%29%5Ex+=+64

2%282%5E%282x%29%29+-+2%5E%282x%29+=+64

2%5E%282x%29+=+64

2%5E%282x%29+=+2%5E6

2x = 6 ------- Bases are equal and so are their exponents

highlight_green%28matrix%281%2C5%2C+x%2C+%22=%22%2C+6%2F2%2C+%22=%22%2C+3%29%29