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) \(\frac{2x}{x-3}\)
\(j^{2x} = k^{3y} = (j^4)(k^2)\)
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
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I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, Indiahttps://www.instagram.com/prepster_education/