Question

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Answer 1

Let 2^a = 3^b = 6^c = k

Thus, we may write:

2 = k^1/a

3 = k^1/b

6 = k^1/c

We know that:

2 × 3 = 6

=> k^1/a × k^1/b = k^1/c

Bases same, powers only:

=> 1/a + 1/b = 1/c

=> (a + b)/ab = 1/c

=> c = ab/(a + b)

hence proved.