SOLUTION: 8^(3x-1) = 64^(x+7) Solve for x

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 8^(3x-1) = 64^(x+7) Solve for x      Log On


   

Question 219277: 8^(3x-1) = 64^(x+7)
Solve for x
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
8%5E%283x-1%29+=+64%5E%28x%2B7%29 Start with the given equation.


8%5E%283x-1%29+=+%288%5E2%29%5E%28x%2B7%29 Rewrite 64 as 8%5E2


8%5E%283x-1%29+=+8%5E%282%28x%2B7%29%29 Multiply the exponents.


3x-1=2%28x%2B7%29 Since the bases are equal, the exponents are equal.


3x-1=2x%2B14 Distribute.


3x=2x%2B14%2B1 Add 1 to both sides.


3x-2x=14%2B1 Subtract 2x from both sides.


x=14%2B1 Combine like terms on the left side.


x=15 Combine like terms on the right side.


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Answer:

So the solution is x=15

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
8^(3x-1) = 64^(x+7)
Solve for x
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Write both sides to the same base:
8^(3x-1) = [8^2]^(x+7)
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Since the bases are equal the exponents are equal:
3x-1 = 2x+14
x = 15
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Cheers,
Stan H.