SOLUTION: I need help with Logarithms. 128^(2x-1)=(1/16)^(x+2)

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Question 288336: I need help with Logarithms.
128^(2x-1)=(1/16)^(x+2)
Found 2 solutions by Alan3354, richwmiller:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
128^(2x-1)=(1/16)^(x+2)
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128 = 2^7 (if you don't know this, you should)
1/16 = 2^-4
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2%5E%287%2A%282x-1%29%29+=+2%5E%28-4%2A%28x%2B2%29%29
Since the bases are the same:
7(2x-1) = -4(x+2)
You can finish it.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
No logarithms needed
128^(2x-1)=(1/16)^(x+2)
128^(2x-1)=(16^(-1))^(x+2)
(2^7)^(2x-1)=2^4(^(-x-2))
2^(14x-7)=2^(-4x-8)
14x-7=-4x-8
18x=-1
x=-1/18