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Exponents

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Exponents

by ng007 » Mon May 30, 2011 4:27 am
If 4^4x = 1600, what is the value of [4^(x-1)]^2?
a)40
b)20
c)10
d)5/2
e)5/4
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by amar66 » Mon May 30, 2011 4:28 am
we have,

4^4x = 1600 ....(Equation 1)

Now lets take a ratio of what we have vs. whats required:

4^4x / [4^(x-1)]^2 = 4 ^ (2x + 2) (Ratio 1)

From Equation 1, 4^2x = 40
Thus 4^(2x+2) = 40 * 16 = 640

So what we need now is 1600/640 (from Ratio 1)
This simplifies to 5/2
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by amar66 » Mon May 30, 2011 4:29 am
OA pls???
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by Anurag@Gurome » Mon May 30, 2011 4:31 am
ng007 wrote:If 4^4x = 1600, what is the value of [4^(x-1)]^2?
a)40
b)20
c)10
d)5/2
e)5/4
4^(4x) = [4^(2x)]^2 = 1600
=> [4^(2x)] = √1600 = 40

And, [4^(x - 1)]^2 = [(4^x)/4]^2 = [4^(2x)]/(4^2) = [4^(2x)]/16 = 40/16 = 5/2

The correct answer is D.
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by ng007 » Mon May 30, 2011 4:34 am
OA is D. thank you.
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