Solution:

Let us take the LHS first:

We can write the given equation as:

9^{1/3 + 1/9 + 1/27 + …∞}

So let us take

$ m\text{ }=\text{ }1/3\text{ }+\text{ }1/9\text{ }+\text{ }1/27\text{ }+\text{ }\ldots \text{ }\infty  $

Where on comparing we have

a = 1/3, r = (1/9) / (1/3) = 1/3

By making use of the formula,

$ {{S}_{\infty }}~=\text{ }a/\left( 1-r \right) $

$ =\text{ }\left( 1/3 \right)\text{ }/\text{ }\left( 1-\left( 1/3 \right) \right) $

$ =\text{ }\left( 1/3 \right)\text{ }/\text{ }\left( \left( 3-1 \right)/3 \right) $

$ =\text{ }\left( 1/3 \right)\text{ }/\text{ }\left( 2/3 \right) $

$ =1/2 $

So, 9^m = 9^1/2 = 3 = RHS

Hence proved.