given that 2^x = 1/square root of 2 and 2^y = 4(square of 2), find exact values of x and y. calculate the exact value of 2^y-x thanks..
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Ok... because sqrt(n) = n^(1/2) and 1/(n^p) = n^(-p) 2^x = sqrt(2)^(-1) = 2^(-1/2) so x = 1/2 2^y=2^2=4 so y = 2 2^(x-y) = 2^(2-1/2) = 2^(3/2) and since n^(3/2) = sqrt(n^3), the final answer is sqrt(8) which is the same as 2*sqrt(2)
2^x = 1/sqrt(2) 2^x = 1/ [2^(1/2)] 2^x = 2^(-1/2) x = -1/2 or -0.5 -----------answer. 2^y = 4sqrt(2) 2^y = (2^2)[2^(1/2)] 2^y = 2^(2 +1/2)) 2^y = 2^(5/2) y = 5/2 or 2.5 -----------answer. 2^(y -x) = 2^(2.5 -(-0.5)) = 2^(3.0) = 8 ---------------answer.
Argh... stupid negatives Bah...
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