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Root/ Exponents...easy way?

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Manager
Manager
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Joined: 14 Dec 2011
Posts: 65
GMAT 1: 630 Q48 V29
GMAT 2: 690 Q48 V37
Followers: 1

Kudos [?]: 24 [0], given: 24

Root/ Exponents...easy way? [#permalink]

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New post 29 Dec 2011, 22:31
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (01:57) correct 35% (02:06) wrong based on 17 sessions

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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
Manager
Manager
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Joined: 17 Oct 2011
Posts: 241
Location: United States
Concentration: Strategy, Marketing
GMAT 1: 720 Q51 V36
Followers: 4

Kudos [?]: 84 [0], given: 36

Re: Root/ Exponents...easy way? [#permalink]

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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.
Senior Manager
Senior Manager
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Joined: 13 May 2011
Posts: 317
WE 1: IT 1 Yr
WE 2: Supply Chain 5 Yrs
Followers: 20

Kudos [?]: 216 [0], given: 11

Re: Root/ Exponents...easy way? [#permalink]

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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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Root/ Exponents...easy way?

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