Maximum value of ${\log }_5(3\mathrm{x}+4\mathrm{y}).$  If ${\mathrm{x}}^2+{\mathrm{y}}^2=25$ is

Since ${\mathrm{x}}^2+{\mathrm{y}}^2=25\Rightarrow \mathrm{x}=5\cos \mathrm{\theta }$ and $\mathrm{y}=5\sin \mathrm{\theta }$

So, therefore ${\log }_5(3\mathrm{x}+4\mathrm{y})={\log }_5(15\cos \mathrm{\theta }+20\sin \mathrm{\theta })$

$range of 15\cos \theta +20\sin \theta =\left[-\sqrt{{15}^2+{20}^2},\sqrt{{15}^2+{20}^2} \right]=\left[-25,25\right]$

$\Rightarrow {\left\{{\log }_5(3\mathrm{x}+4\mathrm{y})\right\}}_{max}={\log }_5 25=2$