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Can someone show how to unsquare exponents?

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Source: — Problem Solving |
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by muzali » Mon Dec 01, 2008 2:27 pm
Square root of a number between 0 and 1 is always greater than the number. Only B [(0.9)^(1/2)] is gretaer than 0.9, hence the answer.
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by aj5105 » Thu May 07, 2009 10:32 pm
Substitute the values for x in the inequality.

(B)
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by mikeCoolBoy » Fri May 08, 2009 2:19 am
I think the best way to attack this problem is to transform the decimal values into powers of base 10.

A) 0.81 ^ 1/2

0.81 is equal to 9^2 x 10 ^ (-2) so ---> 0.81 ^ 1/2 = (9^2 x 10 ^ (-2))^(1/2) = 0.9
we can eliminate A

C) (0.9)^2 = (9x10^(-1))^2 = 9^2 x 10 ^ (-2) = 81 x 10 ^ (-2) = 0.81 we can eliminate C

D) (0.9)(0.99) = (9x10^(-1)) x (99x10^(-2)) = 9 x 99 x 10 ^ (-3) = 801 * 10 ^ (-3) = 0.801
we can eliminate D

E) 1 - (0.01) ^(1/2)
0.01 = 10 ^ (-2) so 1 - ((0.01) ^(1/2)) = 1 - (10 ^ (-2))^(1/2) = 1 - 10 ^ (-1) = 0.9
we can eliminate E

B has to be the answer

B) (0.9)^(1/2) = (9x10^ (-1)^(1/2) = 9 ^(1/2) x 10^ (-1)^(1/2) = 3 x 10^ (-1)^(1/2)

The problem now is to calculate 10^ (-1)^(1/2)
An approximation could be the following

10^ (-1)^(1/2) = (1/10)^(1/2) = 1/(2^(1/2)) x (5^(1/2)) = 1/(1.4 x 2.2) = 1/3.08 = 0.32
so if you multiple 3 x 0.32 = 0.96 which is > 0.9
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