SOLUTION: evaluate e^2ln3 (2^3*2^-5)/2^-7 log4 64 log4 32 log2 4th root(1/8) {{{(3^-2+3^0)^-2}}} log 25 +log 4

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: evaluate e^2ln3 (2^3*2^-5)/2^-7 log4 64 log4 32 log2 4th root(1/8) {{{(3^-2+3^0)^-2}}} log 25 +log 4      Log On


   

Question 66208: evaluate

e^2ln3

(2^3*2^-5)/2^-7

log4 64

log4 32

log2 4th root(1/8)

%283%5E-2%2B3%5E0%29%5E-2

log 25 +log 4
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
evaluate
e^2ln3 = [e^(ln3)]^2 =[3]^2 = 9
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(2^3*2^-5)/2^-7 = 2^-2 / 2^-7 = 2^5 = 32
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log4 64 = log4 (4^3) = 3
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log4 32 = log2 32 / log2 4 = 5/2
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log2 4th root(1/8) = (1/4)log2 [2^-3] = (1/4)(-3) = -3/4
---------------
log 25 +log 4 = log (25*4)= log 100 = 2
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Cheers,
Stan H.