Question

Find a such that
$log_{3}(x) = alog_{5}(x)$
for all values of x > 0 .​

Answer 1

Recall the change-of-base formula:

$\log_5x=\dfrac{\log_3x}{\log_35}$

(you can insert any logarithm on the right side)

It follows that $a=\log_35$.