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UPSC NDA II 2015 Mathematics Paper

Question : 25 of 120

Marks: +1, -0

\log_{8} m + \log_{8} \frac{1}{6} = \frac{2}{3}

m

is equal to

Solution:

Concept:

Use the product rule of logarithms:

\log_a (mn) = \log_a m + \log_a n

Then convert the logarithmic equation to exponential form: if

\log_a x = b

then

x = a^b

Explanation:

Given:

\log_{8} m + \log_{8} \frac{1}{6} = \frac{2}{3}

Apply the product rule:

\log_{8}\left(m \times \frac{1}{6}\right) = \frac{2}{3}

Thus

\log_{8}\left(\frac{m}{6}\right) = \frac{2}{3}

Rewrite in exponential form:

\frac{m}{6} = 8^{2/3}

Simplify

8^{2/3}

8 = 2^3

, so

(2^3)^{2/3} = 2^{3 \times \frac{2}{3}} = 2^2 = 4

Hence

\frac{m}{6} = 4

implies

m = 4 \times 6 = 24

Therefore

m = 24

Answer:

m = 24

, which corresponds to option A.

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