Question

plz ans.................................​

Answer 1

Answer:

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

Then:

2 = k^(1/x)

3 = k^(1/y)

6 = k^(1/z)

We know that:

6 = 2 × 3

=> k^(1/z) = k^(1/x) × k^(1/y)

Bases same, powers only:

=> 1/z = 1/x + 1/y

=> 1/x + 1/y - 1/z = 0

hence proved.

Hope it Helps!!