Answer: Option C
$$\eqalign{
& \,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0 \cr
& \Rightarrow {\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = {7^0} = 1 \cr
& \Rightarrow \sqrt {x + 5} + \sqrt x = {5^1} = 5 \cr
& \Rightarrow {\left( {\sqrt {x + 5} + \sqrt x } \right)^2} = 25 \cr
& \Rightarrow \left( {x + 5} \right) + x + 2\sqrt {x + 5} \sqrt x = 25 \cr
& \Rightarrow 2x + 2\sqrt{ x}\,\, \sqrt {x + 5} = 20 \cr
& \Rightarrow \sqrt {x}\,\, \sqrt {x + 5} = 10 - x \cr
& \Rightarrow x\left( {x + 5} \right) = {\left( {10 - x} \right)^2} \cr
& \Rightarrow {x^2} + 5x = 100 + {x^2} - 20x \cr
& \Rightarrow 25x = 100 \cr
& \Rightarrow x = 4 \cr} $$