Logarithms: Change of Base

$\begin{aligned} \log_5 x + \log_7 x & = \log_{25} x & \small \blue{\text{Converting to same base}} \\ \frac {\log x}{\log 5} + \frac {\log x}{\log 7} & = \frac {\log x}{\log 25} \\ \log x \left(\frac 1{\log 5} + \frac 1{\log 7}\right) & = \frac {\log x}{\log 25} & \small \blue{\text{Since }\frac 1{\log 5} + \frac 1{\log 7} \ne \frac 1{\log 25}} \\ \implies \log x & = 0 \\ x & = \boxed 1 \end{aligned}$

Change of base (I've used natural log here, it doesn't really matter)

$\frac{ln(x)}{ln(5)}$ + $\frac{ln(x)}{ln(7)}$ = $\frac{ln(x)}{ln(25)}$

Moving everything to one side, we get:

$\frac{ln(x)}{ln(5)}$ + $\frac{ln(x)}{ln(7)}$ - $\frac{ln(x)}{ln(25)}$ = 0

Factoring ln(x) out:

ln(x) ( $\frac{1}{ln(5)}$ + $\frac{1}{ln(7)}$ - $\frac{1}{ln(25}$ ) = 0

Since whatever inside the bracket does not equal to 0 (you can further simplify it if you want):

ln(x) = 0

That leads us to:

x = 1