Q.

If \( \frac{\log_{4}x}{2} = \frac{\log_{6}x}{k} = \frac{\log_{9}x}{4} \), find the value of k.

A. 8/3

B. 10/3

C. 3

D. \( \sqrt{3} \)

Find the relation between 4, 6 & 9 and then using it, the relation between \(\log_{x}4, \; \log_{x}6, \; \log_{x}9\) and then replace them with terms related to 2, k & n.

A. 8/3

Approach 1:

4, 6, 9 are in GP

⇒ \(\log_{x}4, \; \log_{x}6, \; \log_{x}9 \) are in AP

⇒ \(\log_{4}x, \; \log_{6}x, \; \log_{9}x \) are in HP

⇒ 2n, kn, 4n are in HP

⇒ 2, k, 4 are in HP

⇒ \( k = \frac{2 \times 2 \times 4}{2 + 4}    \)

Approach 2:

If the given ratios is equal to n, then

\(\log_{4}x=2n \implies \log_{x}4=\frac{1}{2n}\)

Similarly, \(\log_{x}6=\frac{1}{kn}\) and \(\log_{x}9 =\frac{1}{4n}\)

Next, 62 = 4 × 9 \(\implies 2\log_{x}6 = \log_{x}4 + \log_{x}9\)

i.e. \(\frac{2}{kn} = \frac{1}{2n} + \frac{1}{4n} \)

The n gets cancelled from all terms and you should be able to solve it to get value of k.