# Basic logarithm simplification

From $\displaystyle \frac{\log_n b}{\log_n a}$, how do we get $\log_a b$ using algebra?

I haven't been able to do it for about an hour now; I would love some help! Thanks!

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What you are looking at is the "change of base formula" for going from logarithms base $n$ to logarithms base $a$. I'm going to interpret your question as asking why the formula $$\frac{\log_nb}{\log_na} = \log_ab$$ is true.
$$\log_n b = r,\qquad \log_n a = s,\qquad \log_a b = t.$$
That $\log_n b = r$ tells you that $n^r = b$. Similarly, $n^s = a$ (since $\log_n a = s$), and $a^t = b$ (because $\log_ab = t$).
So, $$n^r = b = a^t = (n^s)^t = n^{st}$$ which tells you that $n^r = n^{st}$. But that means that $r=st$. Now let's go back to the expressions for $r$, $s$, and $t$ using logarithms: $r=st$ is the same as $$\log_n b = \bigl( \log_n a\bigr)\bigl(\log_a b\bigr).$$ Now just solve for $\log_ab$.