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difficult Exponential problem

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by Anurag@Gurome » Fri Mar 04, 2011 10:40 am
cuty wrote:If 4^4x = 1600, what is the value of (4x-1)^2?

a. 40
b. 20
c. 10
d. 5/2
e. 5/4
Check the red part.
I think the original expression is [4^(x - 1)]^2

Now, 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 cuty » Fri Mar 04, 2011 11:06 am
yaa....the original expression is [4^(x - 1)]^2

thnks Anurag :)
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by vineeshp » Fri Mar 04, 2011 11:25 am
Anurag@Gurome wrote:
cuty wrote:If 4^4x = 1600, what is the value of (4x-1)^2?

a. 40
b. 20
c. 10
d. 5/2
e. 5/4
Check the red part.
I think the original expression is [4^(x - 1)]^2

Now, 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.
Ya I was wondering, the question seemed to be wrong :D
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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