Logarithm problem

Question

If $a^x=b^y$, then how come $x\log a=y\log b$ holds? Can anyone show me how this is with all steps and necessary logarithm formula?

Answer 1

$\log a^x = \log \overbrace{(a\cdot a \cdots \cdots a)}^x = \log a + \log a + \cdots \cdots + \log a = x\log a$

Since $a^x = b^y\quad $ so $\quad \log a^x = \log b^y \implies x\log a = y\log b$