Exponent problem

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jnellaz: Master | Next Rank: 500 Posts; Posts: 114; Joined: Mon Oct 20, 2008 7:03 pmFollowed by: 5 members; Thanked: 4 times

by jnellaz » Fri Nov 14, 2008 12:53 pm

I was careful to replicate this problem. Please let me know if you have any questions.

((2y^3)^-2/(4x^-5))^-2 =

[spoiler]Answer: 256y^12/x^10
[/spoiler][spoiler]
[/spoiler]

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Source: Beat The GMAT — Problem Solving | Fri Nov 14, 2008 12:53 pm

cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Fri Nov 14, 2008 1:16 pm

Are u sure the 256 is in the numerator with y and not denominator wiht x?

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jnellaz: Master | Next Rank: 500 Posts; Posts: 114; Joined: Mon Oct 20, 2008 7:03 pmFollowed by: 5 members; Thanked: 4 times

by jnellaz » Fri Nov 14, 2008 3:32 pm

Yes. Positive. I double checked.

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Fri Nov 14, 2008 3:52 pm

( (2y^3) ^ - 2/ (4x ^ - 5 ) ) ^ - 2

I have hghlighted the minus signs just to make sure the problem reads as is. Can u please confirm(sorry to bother again

)?

I am getting y ^ 12 / 256 x ^ 10 since the numerator and denominator reciprocates once inside the parenetheses due to negative exponents -2 and -5 and then again reciporocates due to the -2 outside.

I must be missing something here.... Will keep trying

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iamcste: Legendary Member; Posts: 940; Joined: Tue Aug 26, 2008 3:22 amFollowed by: 1 member; Thanked: 55 times

Re: Exponent problem

by iamcste » Fri Nov 14, 2008 4:14 pm

jnellaz wrote:I was careful to replicate this problem. Please let me know if you have any questions.

((2y^3)^-2/(4x^-5))^-2 =

[spoiler]Answer: 256y^12/x^10
[/spoiler][spoiler]
[/spoiler]

Here we go !

Answer and question both are perfect

((2y^3)^-2/(4x^-5))^-2

=( (2^-2*y^-6)/(4x^-5))^-2 ...aplying inner powers..(ab)^m=a^m*b^m

=(( 2^4*y^12)/( 4^-2*x^10)...applying external power..outer brackets..(ab)^m=a^m*b^m

=( ( 2^4* 4^2*y^12) / (x^10)..taking 4^-2 up

=(4^2* 4^2) * (y^12/x^10)....2 ^ 4= 4^2

=( 4 ^4) * (y^12/x^10)

=(256* y^12)/ x^10

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Fri Nov 14, 2008 4:21 pm

Thanks!

I figured out my mistake.

I was taking (2y ) ^ 3 insstead of 2 * (y^3)

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iamcste: Legendary Member; Posts: 940; Joined: Tue Aug 26, 2008 3:22 amFollowed by: 1 member; Thanked: 55 times

by iamcste » Fri Nov 14, 2008 4:28 pm

cramya wrote:Thanks!

I figured out my mistake.

I was taking (2y ) ^ 3 insstead of 2 * (y^3)

No issues...

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jnellaz: Master | Next Rank: 500 Posts; Posts: 114; Joined: Mon Oct 20, 2008 7:03 pmFollowed by: 5 members; Thanked: 4 times

by jnellaz » Sat Nov 15, 2008 7:18 pm

Thanks to you both!

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4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member; Thanked: 27 times; GMAT Score:660(

Re: Exponent problem

by 4meonly » Sun Nov 16, 2008 9:43 am

I still do not understand how u got

iamcste wrote:=( 2^4*y^12)

from

iamcste wrote:=(2^-2*y^-6)

Can somebody clarify?

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iamcste: Legendary Member; Posts: 940; Joined: Tue Aug 26, 2008 3:22 amFollowed by: 1 member; Thanked: 55 times

Re: Exponent problem

by iamcste » Sun Nov 16, 2008 10:37 am

4meonly wrote:I still do not understand how u got

iamcste wrote:=( 2^4*y^12)
from

iamcste wrote:=(2^-2*y^-6)
Can somebody clarify?

(2^-2*y^-6) ^-2

Pls note outside the outer brackets there is a power of -2

Just apply that

(a*b)^m=a^m*b^m

a=2^-2 and m=-2

a^m=(2^-2)^-2

Now apply (a^m)^n=a^mn

mn=-2*-2=4 and a =2

hence a^mn=2^4

similary (y^-6)^-2

=y^12

hence

(2^-2*y^-6) ^-2=( 2^4*y^12)

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4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member; Thanked: 27 times; GMAT Score:660(

by 4meonly » Mon Nov 17, 2008 4:23 am

OK, agree with calculations, but I do not agree with logic.

you made -2 power of numerator. But why you haven't done it with denumerator?

iamcste wrote: =( (2^-2*y^-6)/(4x^-5))^-2 ...aplying inner powers..(ab)^m=a^m*b^m
=(( 2^4*y^12)/( 4^-2*x^10)...applying external power..outer brackets..(ab)^m=a^m*b^m

According to your logic
(ab)/cd = (ab)^2 / cd

but it should be
(ab)/cd = (ab)^2 / (cd)^2

Am I missing something?

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

by parallel_chase » Mon Nov 17, 2008 2:05 pm

Here is a long way of doing this problem, simply for better understanding.

[(2y^3)^-2/(4x^-5)]^-2

Switch the numerator and denominator -power becomes positive

[ (4x^-5) / 2y^3)^-2 ]^2

expand and let it blow 8)

(4^2 * x^-10) / [(2^-2)^2 (y^-6)^2]

=> (4^2 * x^-10) / [(2^-4 (y^-12)]

switch negative powers to numerator and denominator so it becomes positive

(4^2 * 2^4 * y^12) / x^10

256 * y^12 / x^10

Hope this helps.

No rest for the Wicked....

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