SOLUTION: (Root of 256)^(2x +3) = (1/64)^(x-5) (256^1/2)^(2x+3)=(1/64)^(x-5) (256)^(x+1/2)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: (Root of 256)^(2x +3) = (1/64)^(x-5) (256^1/2)^(2x+3)=(1/64)^(x-5) (256)^(x+1/2)      Log On


   

Question 804477: (Root of 256)^(2x +3) = (1/64)^(x-5)
(256^1/2)^(2x+3)=(1/64)^(x-5)
(256)^(x+1/2)
Found 2 solutions by stanbon, erica65404:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(Root of 256)^(2x +3) = (1/64)^(x-5)
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(sqrt(2^8))^(2x+3) = (2^-6)^(x-5)
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(2^4)^(2x+3) = 2^(30-6x)
----
2^(8x+12) = 2^(30-6x)
----
8x+12 = 30-6x
14x = 8
x = 4/7
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Cheers,
Stan H.
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Answer by erica65404(394) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28256%29%5E%282x%2B3%29=%281%2F64%29%5E%28x-5%29
16%5E%282x%2B3%29=%281%2F64%29%5E%28x-5%29
%282%5E4%29%5E%282x%2B3%29=%282%5E%28-6%29%29%5E%28x-5%29
2%5E%284%282x%2B3%29%29=2%5E%28-6%28x-5%29%29
log both sides
4%282x%2B3%29=-6%28x-5%29
8x%2B12=-6x%2B30
14x=18
x=1.29