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Not sure the best way to go about this? If 5^6x=125,000

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Current Student
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26 Jun 2008, 22:10
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If 5^6x=125,000

5^(2x-1)= ??

Thanks!

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Director
Joined: 23 Sep 2007
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Re: Variable Exponent Problem [#permalink]

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26 Jun 2008, 22:28
hardaway7 wrote:

If 5^6x=125,000

5^(2x-1)= ??

Thanks!

do you mean (5^6)(x)= 125,000 ?

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Current Student
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Re: Variable Exponent Problem [#permalink]

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26 Jun 2008, 22:33
Sorry no...
5^(6x) = 125,000

THANKS FOR CHECKING!

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Senior Manager
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Re: Variable Exponent Problem [#permalink]

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26 Jun 2008, 22:43
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)

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Re: Variable Exponent Problem [#permalink]

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26 Jun 2008, 23:22
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

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Re: Variable Exponent Problem [#permalink]

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26 Jun 2008, 23:40
5^(6x) = (5^2x)^3
125000 = (5*10)^3

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

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Re: Variable Exponent Problem [#permalink]

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27 Jun 2008, 00:39
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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.

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Director
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Re: Variable Exponent Problem [#permalink]

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27 Jun 2008, 08:53
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.

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Director
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Re: Variable Exponent Problem [#permalink]

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27 Jun 2008, 12:34
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

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Re: Variable Exponent Problem [#permalink]

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27 Jun 2008, 12:54
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

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Re: Variable Exponent Problem [#permalink]

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27 Jun 2008, 16:39
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.

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Director
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Re: Variable Exponent Problem [#permalink]

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27 Jun 2008, 19:48
and yet, your modesty yielded a wrong answer.

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.

Kudos [?]: 235 [0], given: 0

Director
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Re: Variable Exponent Problem [#permalink]

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27 Jun 2008, 21:23
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...

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Current Student
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Re: Variable Exponent Problem [#permalink]

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09 Jul 2008, 22:54
Just wanted to say thank you to everyone for their help!

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Re: Variable Exponent Problem [#permalink]

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11 Jul 2008, 17:22
gmatnub wrote:
and yet, your modesty yielded a wrong answer.

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.

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Re: Variable Exponent Problem   [#permalink] 11 Jul 2008, 17:22
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Not sure the best way to go about this? If 5^6x=125,000

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