Given $x^2+y^2=25$
$\begin{aligned}
& \Rightarrow \quad y=\sqrt{25-x^2} \\
& \text { Let } z=3 x+4 y \\
& \Rightarrow \quad z=3 x+4 \sqrt{25-x^2}
\end{aligned}$
Differentiating w.r.t. $x$, we get
$\frac{d z}{d x}=3+\frac{4}{2 \sqrt{25-x^2}}(-2 x)$
For maxima put $\frac{d z}{d x}=0$
$\begin{aligned}
& 3=\frac{4 x}{\sqrt{25-x^2}} \\
& 3 \sqrt{25-x^2}=4 x \\
& 9\left(25-x^2\right)=16 x^2 \\
& 9 \times 25-9 x^2=16 x^2 \\
& 9 \times 25=(16+9) x^2 \\
& x^2=9 \\
& \therefore \quad x=3 \\
& \Rightarrow y=\sqrt{25-x^2} \Rightarrow y=4
\end{aligned}$
Now, $\quad z=3 x+4 y$
$=3(3)+4(4)=25$
So, $\log _5[\max (3 x+4 y)]=\log _5[\max (z)]$
$\begin{aligned}
& =\log _5[(25)]=\log _5(5)^2 \\
& =2 \log _5 5=2
\end{aligned}$