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) (Show Source): You can put this solution on YOUR website!
128^(2x-1)=(1/16)^(x+2)
-----------------------
128 = 2^7 (if you don't know this, you should)
1/16 = 2^-4
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Since the bases are the same:
7(2x-1) = -4(x+2)
You can finish it.
Answer by richwmiller(17219) (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
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