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If j, k, x and y are each greater than 1, and
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22 Aug 2022, 07:21
22 Aug 2022, 07:21
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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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If j, k, x and y are each greater than 1, and
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25 Aug 2022, 08:52
25 Aug 2022, 08:52
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GreenlightTestPrep wrote:
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) \(\frac{2x}{x-3}\)
\(j^{2x} = k^{3y} = (j^4)(k^2)\)
Assume \(j = 2\) to simplify things here.
\(j^{2x} = (j^4)(k^2)\)
\(2^{2x} = (2^4)(k^2)\)
Lets try to make the bases equal,
\(2^{2x} = (2^4)(2^2)\)
\(2^{2x} = 2^6\)
i.e. \(2x = 6\) (when the bases are equal, powers are also equal)
\(x = 3\)
This means, \(j = 2\), \(k = 2\), and \(x = 3\)
Now, \(k^{3y} = (j^4)(k^2)\)
\(2^{3y} = (2^4)(2^2)\)
\(2^{3y} = 2^6\)
i.e. i.e. \(3y = 6\) (when the bases are equal, powers are also equal)
\(y = 2\)
We have \(x = 3\) and \(y = 2\)
Plug \(x = 3\) in the option choices and check which option gives \(y = 2\);
A. \(\frac{3x-6}{2x} = \frac{3(3)-6}{2(3)} = \frac{3}{6} = \frac{1}{2}\)
B. \(\frac{2x-6}{3x} = \frac{2(3)-6}{3(3)} = \frac{0}{9} = 0\)
C. \(\frac{2x}{3x-6} = \frac{2(3)}{3(3)-6} = \frac{6}{3} = 2\)
D. \(\frac{3x}{2x-6} = \frac{3(3)}{2(3)-6} = \frac{9}{0}\)
E. \(\frac{2x}{x-3} = \frac{2(3)}{3-3} = \frac{6}{0}\)
Hence, option C
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If j, k, x and y are each greater than 1, and
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31 Aug 2022, 06:18
31 Aug 2022, 06:18
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GreenlightTestPrep wrote:
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) \(\frac{2x}{x-3}\)
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
