Why is $\log_a n = \frac{\log_b n}{\log_b a}$?

Question

What is an intuitive explanation of why $\log_a n = \frac{\log_b n}{\log_b a}$?

Answer 1

Just use the definition :

$$ \log_x(y) = \frac{\text{ln}(y)}{\text{ln}(x)}$$

with ${\text{ln}}$ the "natural" logarithm. So you have :

$$ \frac{\log_b(n)}{\log_b(a)} = \frac{\frac{\text{ln}(n)}{\text{ln}(b)}}{\frac{\text{ln}(a)}{\text{ln}(b)}} = \frac{\text{ln}(n)}{\text{ln}(a)} = \log_a(n) $$

Answer 2

Starting with $a^y=n$ you have $y=\log_a n$

Now if, instead, you took logs (base b) you would have

$\log_b (a^y)=\log_b (n)$

bringing down the power gives

$y\log_b a=\log_b n$

so $y=\frac{\log_b n}{\log_b a}$

so you have $y=\log_a n=\frac{\log_b n}{\log_b a}$

Answer 3

Hint: $$\log_b(n)=\log_b(a^{\log_a(n)})$$

Can you isolate $\log_a(n)$?

Answer 4

Another (weaker, but maybe more intuitive) perspective:

Consider $x, y, z$ all non-zero such that $$ \log_{a}n = x,\qquad \log_{b}n = y,\qquad \log_{b}a = z. $$ These imply that $$ a^{x} = n,\qquad b^{y} = n,\qquad b^{z} = a. $$ Now we have $$ \log_{a}n = \frac{\log_{b}n}{\log_{b}a} \iff x = \frac{y}{z}. $$ It's easy to see that $$ a^{x} = n = b^{y} \iff b^{y/x} = a = b^{z} \iff y/x = z \iff x = y/z. $$