Discussion

Explanation:

Let 3^x = 2^y = 6^z = k

Now, 3x = k ⇒ 3 = k^1/x   …(1)

Similarly, 

2y = k ⇒ 2 = k^1/y   …(2)

6y = k ⇒ 6 = k^1/z   …(3)

We know, 3 × 2 = 6

∴ (1) × (2) = (3)

⇒ k^1/x × k^1/y = k^1/z

⇒ $k^{\frac{1}{x}+\frac{1}{y}} = k^{\frac{1}{z}}$

⇒ $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$

⇒ $z=\frac{xy}{x+y}$

Hence, option (b).

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