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If (2^x)^y = (x^2)^z, the value of y when x is 16 and z is

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SVP
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If (2^x)^y = (x^2)^z, the value of y when x is 16 and z is [#permalink] New post 26 Nov 2007, 02:18
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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?

a) 6
b) 10
c) 16
d) 17
e) 24


Provide explanation when choosing your answer.

regards,
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 [#permalink] New post 26 Nov 2007, 02:38
Expert's post
B

1. (2^16)^y1 = (16^2)^32 ==> 2^(16*y1)=2^(4*2*32) ==> y1=(4*2*32)/16=16

2. (2^8)^y2 = (8^2)^8 ==> 2^(8*y2)=2^(3*2*8) ==> y2=(3*2*8)/8=6

y2-y1=16-6=10
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Re: How much greater is Y? [#permalink] New post 26 Nov 2007, 18:08
tarek99 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?

a) 6
b) 10
c) 16
d) 17
e) 24


Provide explanation when choosing your answer.

regards,



Man enough with the xponent problems jeez :wink:

2^16^y=16^2^32---> 2^16*y=2^4*2*32 We multiply the powers of powers. xample 2^2^2 --> 2^2*2.

So 16y=4*2*32 y= 4*2*2= 16

Next one we get

2^8^y=8^2^8 ---> 2^8y= 2^6*8 8y=6*8 ---> y=6.


So 16-6=10

b.
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 [#permalink] New post 26 Nov 2007, 18:44
dont listen to GMATBLACKBELT, keep the exponents coming ! I need all the practice I can to spot the tricks and solve them in 2 minutes !

oh yeah, got B, 16-6=10
  [#permalink] 26 Nov 2007, 18:44
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If (2^x)^y = (x^2)^z, the value of y when x is 16 and z is

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