JIPMAT 2021 (QA) - If 2^x = 3^y = 6^-z, then (1/x + 1/y + 1/z) is equal to | PYQs + Solutions

JIPMAT 2021

Algebra

Indices

Conceptual

If $2^{x} = 3^{y} = 6^{-z}$ , then $(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is equal to

Correct Option: 1

1) Given:

2^x = 3^y = 6^{-z}

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Let's call this common value

a

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So,

2^x = a

3^y = a

6^{-z} = a

2) Taking

\log_2

of each equality:

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x = \log_2(a)

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y\log_2(3) = \log_2(a)

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-z\log_2(6) = \log_2(a)

3) Therefore:

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\frac{1}{x} = \frac{1}{\log_2(a)}

\frac{1}{y} = \frac{\log_2(3)}{\log_2(a)}

\frac{1}{z} = -\frac{\log_2(6)}{\log_2(a)}

4) Now

\frac{1}{x} + \frac{1}{y} + \frac{1}{z}

= \frac{1}{\log_2(a)} + \frac{\log_2(3)}{\log_2(a)} + -\frac{\log_2(6)}{\log_2(a)}

= \frac{1 + \log_2(3) - \log_2(6)}{\log_2(a)}

= \frac{1 + \log_2(3) - \log_2(2 × 3)}{\log_2(a)}

= \frac{1 + \log_2(3) - (\log_2(2) + \log_2(3))}{\log_2(a)}

= \frac{1 + \log_2(3) - (1 + \log_2(3))}{\log_2(a)}

= 0

Therefore,

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0

The answer is 0.