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!
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!
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.
To see why it is true, remember the meaning of the logarithms. Let's give these quantities names:
$\log_n b = r,\qquad \log_n a = s,\qquad \log_a b = t.$
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 $\log_ab$.