The change-of-base formula

Let $y=\log_b x$. By definition of logarithm, this means $b^y=x$.

By definition of logarithm again (applied to base $c$ this time), $b$ itself can be written as $b=c^{\log_c b}$. Substitute this into $b^y=x$: $$(c^{\log_c b})^y = x \;\Longrightarrow\; c^{y\log_c b} = x,$$ using the power-of-a-power rule $(c^m)^y=c^{my}$.

Take $\log_c$ of both sides. By definition, $c^{y\log_c b}=x$ means exactly $\log_c x = y\log_c b$ (the same injectivity fact $a^u=a^v \Rightarrow u=v$ used in the product-law proof guarantees this step is valid, not just "plausible").

Since $c\ne 1$ we have $\log_c b \ne 0$ (because $c^0=1\ne b$), so we may divide both sides by $\log_c b$: $$y = \dfrac{\log_c x}{\log_c b}.$$ Recalling $y=\log_b x$ gives exactly $\log_b x = \dfrac{\log_c x}{\log_c b}$.