$\begin{aligned} If \log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 0, \\ what is the value of x ? \end{aligned}$ $\eqalign{ & {\text{If}}\,\,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0, \cr & {\text{what}}\,\,{\text{is}}\,\,{\text{the}}\,\,{\text{value}}\,\,{\text{of}}\,\,x? \cr}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 0 \\ \Rightarrow \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 7^{0} = 1 \\ \Rightarrow \sqrt{x + 5} + \sqrt{x} = 5^{1} = 5 \\ \Rightarrow {(\sqrt{x + 5} + \sqrt{x})}^{2} = 25 \\ \Rightarrow (x + 5) + x + 2 \sqrt{x + 5} \sqrt{x} = 25 \\ \Rightarrow 2 x + 2 \sqrt{x} \sqrt{x + 5} = 20 \\ \Rightarrow \sqrt{x} \sqrt{x + 5} = 10 \\ \Rightarrow x (x + 5) = {(10 - x)}^{2} \\ \Rightarrow x^{2} + 5 x = 100 + x^{2} - 20 x \\ \Rightarrow 25 x = 100 \\ \Rightarrow x = 4 \end{aligned}

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