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

Hard

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

✅ Correct Option: 1

Let's call the common value

a

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

Converting to logarithmic form:

From

2^x = a

, we get

x = \log_2(a)

From

3^y = a

, we get

y = \log_3(a)

From

6^{-z} = a

, we get

z = -\log_6(a)

Finding the reciprocals:

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

\dfrac{1}{y} = \dfrac{1}{\log_3(a)}

\dfrac{1}{z} = \dfrac{1}{-\log_6(a)} = -\dfrac{1}{\log_6(a)}

Using the change of base formula to standardize:

\log_3(a) = \dfrac{\log_2(a)}{\log_2(3)}

\log_6(a) = \dfrac{\log_2(a)}{\log_2(6)}

This gives:

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

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

Adding the fractions:

\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{\log_2(a)} + \dfrac{\log_2(3)}{\log_2(a)} - \dfrac{\log_2(6)}{\log_2(a)}= \dfrac{1 + \log_2(3) - \log_2(6)}{\log_2(a)}

Since

\log_2(6) = \log_2(2 \times 3) = \log_2(2) + \log_2(3) = 1 + \log_2(3)

\dfrac{1 + \log_2(3) - \log_2(6)}{\log_2(a)} = \dfrac{1 + \log_2(3) - (1 + \log_2(3))}{\log_2(a)} = \dfrac{0}{\log_2(a)} = 0

\therefore \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0