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Tough PS 2

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New post Updated on: 03 Jun 2009, 18:28
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If \(2401^{2XY} = 49*343^{\frac{2}{3}X}\) and \(Y=3X\)

What is the value of X?

(1) \(0\)

(2) \(\frac{1}{3}\)

(3) \(\frac{1}{2}\)

(4) \(1\)

(5) \(2\)

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Hades


Originally posted by Hades on 03 Jun 2009, 17:10.
Last edited by Hades on 03 Jun 2009, 18:28, edited 1 time in total.
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Re: Tough PS 2  [#permalink]

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New post 03 Jun 2009, 18:23
Hades wrote:
If \(2401^{2XY} = 49*243^{\frac{2}{3}X}\) and \(Y=3X\)

What is the value of X?

(1) \(0\)
(2) \(\frac{1}{3}\)
(3) \(\frac{1}{2}\)
(4) \(1\)
(5) \(2\)


\(2401^({2XY}) = 49*243^({\frac{2}{3}X})\)
\(7^({4*2X*3x}) = 7^{2} * 3^({\frac{5*2}{3}X})\)
\(7^({24X^{2} - 2}) = 3^({\frac{10}{3}X})\)

What next ????????
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Re: Tough PS 2  [#permalink]

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New post 03 Jun 2009, 18:27
omfg It should be 343, sorry
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Re: Tough PS 2  [#permalink]

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New post 03 Jun 2009, 18:30
Also

\(2401^{2XY}\)
\(=49^{4XY}\)
\(=7^{8XY}\)
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Re: Tough PS 2  [#permalink]

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New post 03 Jun 2009, 18:39
Hades wrote:
If \(2401^{2XY} = 49*343^{\frac{2}{3}X}\) and \(Y=3X\)

What is the value of X?

(1) \(0\)
(2) \(\frac{1}{3}\)
(3) \(\frac{1}{2}\)
(4) \(1\)
(5) \(2\)


\(2401^{2XY} = 49*343^{\frac{2}{3}X}\)
\(7^{4*2X*3x} = 7^{2} * 7^{\frac{3*2}{3}X}\)
\(7^{24X^{2}} = 7^{{4X}\)
\(24X^{2} = 4X\)
\(6X^{2} - X = 0\)
\(x (6X-1) = 0\)
\(X = 0 or \frac{1}{6}\)
\(So X = 0\)
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Re: Tough PS 2  [#permalink]

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New post 03 Jun 2009, 19:22
1
GMAT TIGER, check your exponent algebra...


\(2401^{2XY} = 49*343^{\frac{2}{3}X}\)

\(49^{4XY} = 7^{2}(7^{3})^{\frac{2}{3}X}\)

\(7^{8XY} = 7^{2}7^{\frac{3*2}{3}X}\)

\(7^{8XY} = 7^{2}7^{2X}\)

\(7^{8XY} = 7^{2+2X}\)

\(\longrightarrow 8XY=2+2X\)



Substitute in \(Y=3X\)



\(8X(3X)=2+2X\)

\(24X^{2}=2+2X\)

\(12X^{2}=1+X\)

\(12X^{2}=1+X\)

You can either factor:

\(12X^{2} - X - 1=0\)

\(X^{2} - \frac{1}{12X} - \frac{1}{12} = 0\)

\(X^{2} + \frac{1}{4X} - \frac{1}{3X} - \frac{1}{12} = 0\)

\(X(X + \frac{1}{4}) - \frac{1}{3}(X + \frac{1}{4}) = 0\)

\((X - \frac{1}{3})(X + \frac{1}{4}) = 0\)

\(\longrightarrow X=\frac{1}{3}\) OR \(X=-\frac{1}{4}\)


Or substitute:
(1) \(X=0 \longrightarrow 0 = 1\)

(2) \(X=\frac{1}{3} \longrightarrow 4/3 = 4/3\)

(3) \(X=\frac{1}{2} \longrightarrow 3 = 3/2\)

(4) \(X=1 \longrightarrow 12 = 2\)

(5) \(X=2 \longrightarrow 48 = 3\)


Final Answer, 2: \(X= \frac{1}{3}\).
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Re: Tough PS 2  [#permalink]

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New post 03 Jun 2009, 23:47
Hades wrote:
If \(2401^{2XY} = 49*343^{\frac{2}{3}X}\) and \(Y=3X\)

What is the value of X?

(1) \(0\)
(2) \(\frac{1}{3}\)
(3) \(\frac{1}{2}\)
(4) \(1\)
(5) \(2\)


\(2401^{2XY} = 49*343^{\frac{2}{3}X}\)
\(7^{4*2X*3x} = 7^{2} * 7^{\frac{3*2}{3}X}\)
\(7^{24X^{2}} = 7^{2X+2}\)
\(24X^{2} = 2x+2\)
\(12X^{2} - x - 1 = 0\)
\((x-3) (x+4) = 0\)
\(X = {-4} or \frac{1}{3}\)
\(So X = \frac{1}{3}\)


Agree that I did multiplication instead of addition.
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Re: Tough PS 2  [#permalink]

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New post 04 Jun 2009, 00:06
GMAT TIGER wrote:
Hades wrote:
If \(2401^{2XY} = 49*343^{\frac{2}{3}X}\) and \(Y=3X\)

What is the value of X?

(1) \(0\)
(2) \(\frac{1}{3}\)
(3) \(\frac{1}{2}\)
(4) \(1\)
(5) \(2\)


\(2401^{2XY} = 49*343^{\frac{2}{3}X}\)
\(7^{4*2X*3x} = 7^{2} * 7^{\frac{3*2}{3}X}\)
\(7^{24X^{2}} = 7^{2X+2}\)
\(24X^{2} = 2x+2\)
\(12X^{2} - x - 1 = 0\)
\((x-3) (x+4) = 0\)
\(X = \frac{-1}{4} or \frac{1}{3}\)
\(So X = \frac{1}{3}\)


Agree that I did multiplication instead of addition.


\((x-3) (x+4) = 0\)
\(\longrightarrow x=3\) OR \(x=-4\)....

unless you meant 1 over
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Re: Tough PS 2  [#permalink]

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New post 04 Jun 2009, 07:07
Hades wrote:
GMAT TIGER wrote:
Hades wrote:
If \(2401^{2XY} = 49*343^{\frac{2}{3}X}\) and \(Y=3X\)

What is the value of X?

(1) \(0\)
(2) \(\frac{1}{3}\)
(3) \(\frac{1}{2}\)
(4) \(1\)
(5) \(2\)


\(2401^{2XY} = 49*343^{\frac{2}{3}X}\)
\(7^{4*2X*3x} = 7^{2} * 7^{\frac{3*2}{3}X}\)
\(7^{24X^{2}} = 7^{2X+2}\)
\(24X^{2} = 2x+2\)
\(12X^{2} - x - 1 = 0\)
\((x-3) (x+4) = 0\)
\(X = {-4} or \frac{1}{3}\)
\(So X = \frac{1}{3}\)


Agree that I did multiplication instead of addition.


\((x-3) (x+4) = 0\)
\(\longrightarrow x=3\) OR \(x=-4\)....

unless you meant 1 over


Obviously not.
Don't worry. Thats typo. :oops:
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Re: Tough PS 2  [#permalink]

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New post 04 Jun 2009, 11:47
2401^(2xy) = 49*343^(2/3x)
2401^(2xy) = 49*49^x
2401^(2xy) = 2401^x
(2xy)=x
y=1/2

y=3x
x=1/6
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New post 04 Jun 2009, 12:59
amolsk11 wrote:
2401^(2xy) = 49*343^(2/3x)
2401^(2xy) = 49*49^x
2401^(2xy) = 2401^x
(2xy)=x
y=1/2

y=3x
x=1/6


\(49(49^{x}) \neq 2401^x\)
...
\(49(49^{x}) = 49^{x+1}\)
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Re: Tough PS 2  [#permalink]

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New post 24 Jun 2009, 11:50
I think what GMAT TIGER meant was

12x^2 -x - 1 = 0
(3x -1)(4x +1) = 0
x = 1/3 or -1/4

I think this is a bit easier to deal with than deal with the denominator. Great question overall though.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Tough PS 2 &nbs [#permalink] 24 Jun 2009, 11:50
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