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Maximum value of $\log_{5} (3 x + 4 y),$ if $x^{2} + y^{2} = 25$ is

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Detailed Solution

Since $x^{2} + y^{2} = 25 \Rightarrow x = 5 \cos θ a n d y = 5 \sin θ$ So, therefore,

$\log_{5} (3 x + 4 y) = \log_{5} (15 \cos θ + 20 \sin θ)$

$\Rightarrow {\{\log_{5} (3 x + 4 y)\}}_{\max} = 2$

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