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![If j, k, x and y are each greater than 1, and [m]j^{2x} = k^{3y} = j^4](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If j, k, x and y are each greater than 1, and [m]j^{2x} = k^{3y} = j^4"){kind=icon}
22 Aug 2022, 08:20
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C
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Difficulty:
65%
(hard)
Question Stats:
65% (03:11) correct
35%
(02:57)
wrong
based on 239
sessions
History
Date
Time
Result
Not Attempted Yet
If \(j\), \(k\), \(x\) and \(y\) are each greater than \(1\), and \(j^{2x} = k^{3y} = j^4k^2\), then what is the value of \(y\) in terms of \(x\)?
A) \(\frac{3x-6}{2x}\)
B) \(\frac{2x-6}{3x}\)
C) \(\frac{2x}{3x-6}\)
D) \(\frac{3x}{2x-6}\)
E) \(\frac{2x}{x-3}\)
A) \(\frac{3x-6}{2x}\)
B) \(\frac{2x-6}{3x}\)
C) \(\frac{2x}{3x-6}\)
D) \(\frac{3x}{2x-6}\)
E) \(\frac{2x}{x-3}\)
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23 Aug 2022, 07:31
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BrentGMATPrepNow
No one wants to try this one?
Strategy: Make one of the expressions with look like another expression.
Take: \(j^{2x} = j^4k^2\)
Divide both sides of the equation by \(j^4\) to get: \(\frac{j^{2x}}{j^4} = \frac{j^4k^2}{j^4}\)
Simplify: \(j^{2x - 4} = k^2\)
Raise both sides of the equation to the power of \(1.5y\) to get: \((j^{2x - 4})^{1.5y} = (k^2)^{1.5y}\)
Simplify: \(j^{3xy - 6y} = k^{3y}\)
It's given that \(j^{2x} = k^{3y}\), we can replace \(k^{3y}\) with \(j^{2x}\) to get: \(j^{3xy - 6y} = j^{2x}\)
Since the bases are equal, we can conclude the exponents are equal to get: \(3xy - 6y = 2x\)
Factor the left side: \(y(3x - 6) = 2x\)
Solve for y: \(y = \frac{2x}{3x - 6}\)
Answer: C
General Discussion
23 Aug 2022, 09:14
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BrentGMATPrepNow
BrentGMATPrepNow, let me try it with slightly different method.
\(j^{2x} = k^{3y} = j^4k^2\)
\(j^{2x} * k^{3y} = j^4k^2*j^4k^2=j^8*k^4\)
Let us equate powers of j and k.
\(2x=8…..x=4\)
\(3y=4……y=\frac{4}{3}\)
Now, substitute x as 4 and check for option that gives 4/3 as the answer.
A) \(\frac{3x-6}{2x}=\frac{3*4-6}{2*4}=\frac{3}{4}\)…NO
B) \(\frac{2x-6}{3x}=\frac{2*4-6}{3*4}\)…NO
C) \(\frac{2x}{3x-6}=\frac{2*4}{3*4-6}=\frac{8}{6}\)…YES
D) \(\frac{3x}{2x-6} =\frac{3*4}{2*4-6}=\frac{12}{2}\)….NO
E) \(\frac{2x}{x-3} =\frac{2*4}{4-3}=\frac{8}{1}\)…NO
OR
Take j as 2 and k as 4…….I have taken j as factor of k.
\(2^{2x} = 4^{3y} = 2^44^2\)
\(2^{2x}=2^44^2=2^8……..2x=8……x=4\)
\(4^{3y}=2^{6y}=2^44^2=2^8……..6y=8……y=4/3\)
Now, substitute x as 4 and check for option that gives 4/3 as the answer.
SAME as above
C
