Determine the value of 'a' which satisfies the equation $$9^{\sqrt{a}}+40^{\sqrt{a}}=41^{\sqrt{a}}$$

Expression : $$9^{\sqrt{a}}+40^{\sqrt{a}}=41^{\sqrt{a}}$$

Since equation consists of natural numbers, then 'a' must be an integer. Thus, substituting 'a' by perfect square numbers :

Put $$a=1$$

=> $$9^{\sqrt{1}}+40^{\sqrt{1}}=41^{\sqrt{1}}$$

L.H.S. = $$9+40=49\neq$$ R.H.S.

Now, putting $$a=4$$

=> $$9^{\sqrt{4}}+40^{\sqrt{4}}=41^{\sqrt{4}}$$

L.H.S. = $$9^2+40^2=81+1600=1681$$

R.H.S. = $$41^2=1681$$

Thus, $$a=4$$

=> Ans - (D)