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Current Student
Joined: 03 Dec 2007
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If 5^6x=125,000

5^(2x-1)= ??

Thanks!
Director
Joined: 23 Sep 2007
Posts: 797
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Re: Variable Exponent Problem [#permalink]  26 Jun 2008, 21:28
hardaway7 wrote:

If 5^6x=125,000

5^(2x-1)= ??

Thanks!

do you mean (5^6)(x)= 125,000 ?
Current Student
Joined: 03 Dec 2007
Posts: 84
Schools: NYU STERN
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Kudos [?]: 6 [0], given: 0

Re: Variable Exponent Problem [#permalink]  26 Jun 2008, 21:33
Sorry no...
5^(6x) = 125,000

THANKS FOR CHECKING!
Senior Manager
Joined: 06 Jul 2007
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Re: Variable Exponent Problem [#permalink]  26 Jun 2008, 21:43
hardaway7 wrote:

If 5^6x=125,000

5^(2x-1)= ??
Thanks!

5^6x = (5^2x)^3=12500

5^2x = (12500)^(1/3) = 5*[(100)^(1/3)]

so 5^(2x-1) = [5^(2x)]/5 = (100)^(1/3)
Current Student
Joined: 03 Dec 2007
Posts: 84
Schools: NYU STERN
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Re: Variable Exponent Problem [#permalink]  26 Jun 2008, 22:22
thanks but sadly I am still confused! I am having trouble understanding what exactly you did to the 125,000 (i believe it was a mistake you left off a zero but I am confused aside from that)

I get why you did 5^[(2x)3] but am have trouble breaking out the other side!

Thanks so much for your help
SVP
Joined: 17 Jun 2008
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Re: Variable Exponent Problem [#permalink]  26 Jun 2008, 22:40
5^(6x) = (5^2x)^3
125000 = (5*10)^3

Hence, 5^2x = 50
and 5^(2x-1) = 50/5 = 10
Senior Manager
Joined: 06 Jul 2007
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Re: Variable Exponent Problem [#permalink]  26 Jun 2008, 23:39
1
KUDOS
hardaway7 wrote:
thanks but sadly I am still confused! I am having trouble understanding what exactly you did to the 125,000 (i believe it was a mistake you left off a zero but I am confused aside from that)

I get why you did 5^[(2x)3] but am have trouble breaking out the other side!

Thanks so much for your help

5^6x = (5^2x)^3=12500

so, now, you will have to take the 1/3rd power of both sides.

(5^6x)^(1/3) = (12500)^(1/3)

(5^6x)^(1/3) = 5^(2x)
and
12500 = 125*100 = (5^3)*(100)

so (12500)^(1/3) = [(5^3)^(1/3)]*(100)^(1/3)
= 5*[(100)^(1/3)]

so 5^(2x) = 5*[(100)^(1/3)]

now, 5^(2x-1) = (5^2x)/5

substituting the value of 5^2x in the above equation :

5^(2x-1) = {5*[(100)^(1/3)]}/5 = 100^(1/3)

please let me know if u r still not clear.
Director
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Re: Variable Exponent Problem [#permalink]  27 Jun 2008, 07:53
scthakur wrote:
5^(6x) = (5^2x)^3
125000 = (5*10)^3

Hence, 5^2x = 50
and 5^(2x-1) = 50/5 = 10

interesting, this seems to be correct too.

nevermind, the other solution left off a zero.
Director
Joined: 27 May 2008
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Re: Variable Exponent Problem [#permalink]  27 Jun 2008, 11:34
hardaway7 wrote:

If 5^6x=125,000

5^(2x-1)= ??

Thanks!

This is how I'll approach

"5^6x=125,000" looks to me simpler than "5^(2x-1)= " So I'll try to simplify "5^(2x-1)= "

Y = 5^(2x-1) = (5^2x)/5
Y^3 = (5^6x) / 125
= 125000/125
= 1000

So Y = (1000)^1/3 = 10
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
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Re: Variable Exponent Problem [#permalink]  27 Jun 2008, 11:54
hardaway7 wrote:
Sorry no...
5^(6x) = 125,000

THANKS FOR CHECKING!

NEVER MIND..I figured it out..shit..5^6x=5^(2x)^3....not 5^3 * 5^(2x)...duh..what a stupid mistake i made..and even wrote it out on my response..

what am i doing wrong??

5^(6x)=5^(2x)^3=125,000

5^(2x)=125,000/125=1000
5^(2x-1)=1000/5...=200
SVP
Joined: 21 Jul 2006
Posts: 1551
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Kudos [?]: 305 [0], given: 1

Re: Variable Exponent Problem [#permalink]  27 Jun 2008, 15:39
believe it or not guys, i just solved this problem off the top of my head within 30 seconds without even using a scratch paper!!

ok,
5^6x=125,000

5^(2x-1)=?

we know that if we find the cube root of (5^6x), we will have 5^2x. If you had memorized your cubes, you would have known that 5^3 is equal to 125. So you automatically know that the cube root of 125,000 is 5,000. Also, 5^(2x-1) means 5^2x / 5, therefore: 5000 / 5 = 1000 is our answer.
Director
Joined: 23 Sep 2007
Posts: 797
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Kudos [?]: 100 [0], given: 0

Re: Variable Exponent Problem [#permalink]  27 Jun 2008, 18:48

tarek99 wrote:
believe it or not guys, i just solved this problem off the top of my head within 30 seconds without even using a scratch paper!!

ok,
5^6x=125,000

5^(2x-1)=?

we know that if we find the cube root of (5^6x), we will have 5^2x. If you had memorized your cubes, you would have known that 5^3 is equal to 125. So you automatically know that the cube root of 125,000 is 5,000. Also, 5^(2x-1) means 5^2x / 5, therefore: 5000 / 5 = 1000 is our answer.
Director
Joined: 27 May 2008
Posts: 550
Followers: 7

Kudos [?]: 209 [0], given: 0

Re: Variable Exponent Problem [#permalink]  27 Jun 2008, 20:23
tarek99 wrote:
the cube root of 125,000 is 5,000.

here is what went wrong.

I do these mistakes all the time, on an average 3 - 5 times in a test. and I'm hoping on the GMAT day, all such errors will be in experimental questions...
Current Student
Joined: 03 Dec 2007
Posts: 84
Schools: NYU STERN
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Kudos [?]: 6 [0], given: 0

Re: Variable Exponent Problem [#permalink]  09 Jul 2008, 21:54
Just wanted to say thank you to everyone for their help!
SVP
Joined: 21 Jul 2006
Posts: 1551
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Kudos [?]: 305 [0], given: 1

Re: Variable Exponent Problem [#permalink]  11 Jul 2008, 16:22
gmatnub wrote:

tarek99 wrote:
believe it or not guys, i just solved this problem off the top of my head within 30 seconds without even using a scratch paper!!

ok,
5^6x=125,000

5^(2x-1)=?

we know that if we find the cube root of (5^6x), we will have 5^2x. If you had memorized your cubes, you would have known that 5^3 is equal to 125. So you automatically know that the cube root of 125,000 is 5,000. Also, 5^(2x-1) means 5^2x / 5, therefore: 5000 / 5 = 1000 is our answer.

Sorry! the cube root of 125,000 is simply 50! so when you divide 50 /5, you will have 10.
Re: Variable Exponent Problem   [#permalink] 11 Jul 2008, 16:22
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