Logarithm with circle

By Cauchy-Schwarz, we have $(3x+4y)^2\leq (3^2+4^2)(x^2+y^2)=25^2$ , hence $3x+4y\leq 25$ and so $\log_5(3x+4y)\leq\log_5(25)=\boxed{2}.$

Since $x^2+y^2=25$ , we can substitute $x=5\cos{\theta}$ and $y = 5\sin{\theta}$ . Since $\log_5{(3x+4y)} \ge 0$ , it has maximum value when $3x+4y$ is maximum or $f(\theta) = 15 \cos{\theta}+20\sin{\theta}$ is maximum.

We note that:

$\begin{aligned} f(\theta) & = 15 \cos{\theta}+20\sin{\theta} = 25 \left( \frac{3}{5}\cos{\theta}+\frac{4}{5} \sin{\theta} \right) \\ & = 25 \left( \sin{\phi}\cos{\theta}+\cos{\phi }\sin{\theta} \right) \end{aligned}$

$\quad \quad \space = 25 \sin{(\theta + \phi)}$ , where $\phi = \tan^{-1} {\frac{3}{4}}$ .

$\Rightarrow f(\theta)_{max} = 25$ , when $\sin{(\theta + \phi)} = 1$ , and maximum $\log_5{(3x+4y)} = \log_5 {25} = \boxed{2}$

Let x = 5cos@ and y=5sin@ . Maximum value of 3x+4y = maximum value of 15cos@+20sin@. = $\sqrt{15^{2}+20^{2}}$ =25. Hence $log_{5}25 =2$