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Exponents--Not hard, but stuck...

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by NDroho » Sat Jul 31, 2010 10:21 am
I'm having trouble grasping the concept of this exponent problem. I got the right answer, but just don't get it. The "detailed" explanation doesn't seem to help either. I have put in RED the correct answer, as well as as the explanation given (as I said, I'm not following how they explained it). Here is the question:

If X^64=64, then x^32=?
A) 8 or -8
B) 12 or -12
C) 16
D) 32 or -32
E) 48

Explanation:
X^64=(X^32)^2
X^64=64
(X^32)^2=64
Since the only two numbers whose square is 64 are 8 and -8, X^32=8 or X^32=-8
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Source: — Problem Solving |

by clock60 » Sat Jul 31, 2010 10:58 am
x^64=x^(32*2)=(x^32)^2
64=8^2

(x^32)^2=8^2
(x^32)^2-8^2=0, here apply a^2-b^2=(a-b)(a+b)
(x^32-8)(x^32+8)=0
x^32=8
x^32=-8
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by pradeepkaushal9518 » Sat Jul 31, 2010 11:01 am
what to do in this question?
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by dinesh19aug » Sat Jul 31, 2010 2:16 pm
NDroho wrote:I'm having trouble grasping the concept of this exponent problem. I got the right answer, but just don't get it. The "detailed" explanation doesn't seem to help either. I have put in RED the correct answer, as well as as the explanation given (as I said, I'm not following how they explained it). Here is the question:

If X^64=64, then x^32=?
A) 8 or -8
B) 12 or -12
C) 16
D) 32 or -32
E) 48

Explanation:
X^64=(X^32)^2
X^64=64
(X^32)^2=64
Since the only two numbers whose square is 64 are 8 and -8, X^32=8 or X^32=-8
Another way to visualize this question:

Assume x^32 = a
So X^64 ===> (x^32^2) === a^2

So given that a ^2 = 64
a = +8 or -8
==> Replacing a with x^32 = + 8 or -8

It's a very nice question. In exponents All you need to focus on is to keep the bases same. Rest is easy.
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by KrazyKarl » Sat Jul 31, 2010 4:04 pm
Thanks Dinesh - that's a pretty slick way to solve that!
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