If $x^{2} + y^{2} = 25,$ then the maximum value of ${log}_{5} | 3 x + 4 y |$ is

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Solution

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

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