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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- Oct 1st 2005, 03:08 AMx-disturbed-xbrain teaser
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.. - Oct 1st 2005, 05:47 AMWillyG
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) - Oct 1st 2005, 10:04 AMticbol
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. - Oct 1st 2005, 10:08 AMWillyG
Argh... stupid negatives :mad:

Bah...