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Root/ Exponents...easy way?

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Root/ Exponents...easy way? [#permalink] New post 29 Dec 2011, 22:31
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Question Stats:

71% (01:12) correct 28% (02:40) wrong based on 7 sessions
If (2^x)^y = (x^2)^z, the value of y when x is 16 and z is 32 is how much greater than the value of y when x and z are both 8 ?

6
10
16
17
24


My Idea

2^(x*y) = x^(2*z)
2^(16y) = 16^64 ---- > no idea how to get y out of this
2^(8y) = 8^16 ---- > and again no idea how to get y

How to I solve those equations?

Thanks a lot!
[Reveal] Spoiler: OA
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Re: Root/ Exponents...easy way? [#permalink] New post 29 Dec 2011, 23:14
Impenetrable wrote:
If (2^x)^y = (x^2)^z, the value of y when x is 16 and z is 32 is how much greater than the value of y when x and z are both 8 ?

6
10
16
17
24


My Idea

2^(x*y) = x^(2*z)
2^(16y) = 16^64 ---- > no idea how to get y out of this
2^(8y) = 8^16 ---- > and again no idea how to get y

How to I solve those equations?

Thanks a lot!


2^(xy)=x^(2z)
2^(16y)=16^64
but,16=2^4
so, 2^(16y)=2^(4*64)
16y=4*64
y=16

similarly,
8=2^3
so, 8y=3*2*8
or, y=6
Answer: B

Hope this helps.
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Re: Root/ Exponents...easy way? [#permalink] New post 30 Dec 2011, 03:01
for the 1st condition: 2^16y=2^(16*16)
y=16
for the 2nd condition: 2^8y=2^(6*8)
y=6
16-6=10=B
Re: Root/ Exponents...easy way?   [#permalink] 30 Dec 2011, 03:01
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