Exponents and Estimating

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aznmexicana: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Sat Aug 15, 2009 5:10 pm; Thanked: 1 times

by aznmexicana » Tue Jul 20, 2010 10:24 pm

Where would one start in solving this??

x^4+ y^4=100 then what is the greatest possible value of x between?

a)0-3 b) 3-6 c) 6-9 d) 9-12 e) 12-15[/list]

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Source: Beat The GMAT — Problem Solving | Tue Jul 20, 2010 10:24 pm

surajgarg: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Mon May 10, 2010 11:55 pm; Thanked: 4 times

by surajgarg » Tue Jul 20, 2010 10:51 pm

aznmexicana wrote:Where would one start in solving this??

x^4+ y^4=100 then what is the greatest possible value of x between?

a)0-3 b) 3-6 c) 6-9 d) 9-12 e) 12-15[/list]

Its asking for greatest possible value of x.

This is possible when y is least.

Since the exponent of y is 4, an even number, then y^4 will be positive, so we can fix y = 0.

That leaves x^4 = 100
i.e. x^2 = 10
i.e. x = 3.16

So answer is B.

Last edited by surajgarg on Wed Jul 21, 2010 6:28 am, edited 1 time in total.

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taneja.niks: Senior | Next Rank: 100 Posts; Posts: 70; Joined: Tue Jun 29, 2010 9:41 am; Thanked: 2 times

by taneja.niks » Wed Jul 21, 2010 5:13 am

Since the exponent of y is 4, an even number, then we can fix y = 0

what makes u say this statement tht exponent is 4 so its a even number 3^4 is 81 is it a even number????

Please post the answer for the same........

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surajgarg: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Mon May 10, 2010 11:55 pm; Thanked: 4 times

by surajgarg » Wed Jul 21, 2010 6:28 am

taneja.niks wrote:Since the exponent of y is 4, an even number, then we can fix y = 0

what makes u say this statement tht exponent is 4 so its a even number 3^4 is 81 is it a even number????

Please post the answer for the same........

Hey niks,

Thanks for the observation. I missed adding a phrase there. I have corrected my post above.

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VS2013: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Sun Jul 11, 2010 6:52 pm; Location: Washington, DC; Thanked: 2 times; GMAT Score:630

by VS2013 » Thu Jul 22, 2010 4:17 pm

For the greatest possible value of x, y^4 must be 0.
Therefore, x^4=100= 10*10
Since 3*3=9, then x is slightly greater than 3.
Therefore, answer is b, 3-6.

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zareentaj: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Fri Aug 06, 2010 11:12 pm

by zareentaj » Wed Aug 11, 2010 4:26 am

VS2013 wrote:For the greatest possible value of x, y^4 must be 0.
Therefore, x^4=100= 10*10
Since 3*3=9, then x is slightly greater than 3.
Therefore, answer is b, 3-6.

Thank you VS2013.

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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Wed Aug 11, 2010 7:33 am

aznmexicana wrote:Where would one start in solving this??

x^4+ y^4=100 then what is the greatest possible value of x between?

a)0-3 b) 3-6 c) 6-9 d) 9-12 e) 12-15[/list]

x^4+ y^4=100

now
x^4+ y^4
= x^4+ y^4 +4(x^2)(y^2) - 4(x^2)(y^2)
=(x^2 +y^2)^2 - (2xy)^2
=(x^2 +y^2 +2xy) (x^2 +y^2 -2xy)
=(x+y)^2 * (x-y)^2
=10^2

so (x+y)(x-y) =10

x^2 -y^2 = 10

x^2 = 10 + y^2

which would mean
x^2 > 10

x > root10
ie. x > 3.16
so b fits....

This seems to be a lenghty way... but I have shown all the steps here.. to make things

people seeing a x^4+ y^4 at once jump to (x+y)^2 * (x-y)^2

the idea in exponent problems always is to try and reduce powers to as manageable as possible...

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