Question

If $x^2+y^2=25,$ then the maximum value of ${\log }_5|3x+4y|$ is

Answer 1

$x=5\cos \theta ,y=5\sin \theta$
${\log }_5|3x+4y|={\log }_5|15\cos \theta +20\sin \theta |$
$={\log }_{5}5|3\cos \theta +4\sin \theta |$
$=1+{\log }_5|3\cos \theta +4\sin \theta |$
$\because 3\cos \theta +4\sin \theta \in [-\sqrt{3^2+4^2},\sqrt{3^2+4^2}]$ i.e., $[-5,5]$
$\Rightarrow |3\cos \theta +4\sin \theta |\in [0,5]$
So, the maximum value of ${\log }_5|3x+4y|=1+{\log }_{5}5=2$