Rooted logs

We need to state whether ${log_{10}(\sqrt{15}-\sqrt{14})} {<} {log_{0.1}(\sqrt{16}-\sqrt{15})}$ is true.

Using property of logarithms,

${log_{10}(\sqrt{15}-\sqrt{14})} {<} {log_{10}(\frac{1}{\sqrt{16}-\sqrt{15})})}$

Rationalising,

${log_{10}(\sqrt{15}-\sqrt{14})} {<} {log_{10}(\sqrt{16}+\sqrt{15})}$

This statement is obviously true as $\log _{ 10 }{ x }$ is a monotonically increasing function.

$\begin{cases} \sqrt{15}-\sqrt{14} < 1 & \log_{\color{#D61F06}{10}}{\sqrt{15}-\sqrt{14}} < 0 \\ \sqrt{16}-\sqrt{15} < 1 & \log_{\color{#D61F06}{10}}{\sqrt{16}-\sqrt{15}} < 0 \end{cases}$

But $\log_{\color{#3D99F6}{0.1}}{\sqrt{16}-\sqrt{15}} = \dfrac {\log_{\color{#D61F06}{10}}{\sqrt{16}-\sqrt{15}}}{\log_{\color{#D61F06}{10}}{\color{#3D99F6}{0.1}}} = \dfrac {\log_{\color{#D61F06}{10}}{\sqrt{16}-\sqrt{15}}}{\color{#3D99F6}{-1 }} \color{#3D99F6}{> 0}$

$\Rightarrow \log_{\color{#D61F06}{10}}{\sqrt{15}-\sqrt{14}} < \log_{\color{#3D99F6}{0.1}}{\sqrt{16}-\sqrt{15}} \space$ is $\space \boxed{\color{#3D99F6}{True}}$

Using graph of logarithmic function for base <1 & base>1 we can see that argument of logarithm in both cases lies in (0,1) and on LHS base >1 => LHS <0 i.e negative while on RHS ,base <1 => RHS >0 i.e positive.