One of the rules of logarithms that I've been taught is:
$\log_b(x)$ is equivalent to $\dfrac{\log_k(x)}{\log_k(b)}$.
Recently I've also seen another rule that says:
$\log_b(x)$ is equivalent to $\log_b(k)\log_k(x)$.
Are these equivalent (perhaps via some refactoring) ?
Best Answer
First: They are not "equivalent", they are equal.
That said: remember the meaning of "$\log_b(a) = r$". It means that $r$ is the exponent to which you must raise $b$ in order to get $a$; that is, $\log_b(a)=r$ is equivalent to $b^r=a$.
So... why is $\log_b(x)=\frac{\log_k(x)}{\log_k(b)}$? Because if you let $r=\log_b(x)$, $s=\log_k(x)$ and $t=\log_k(b)$, then that means that $b^r = x$, $k^s = x$, and $k^t=b$.
Therefore,
$$k^s = x = b^r = (k^t)^r = k^{tr}.$$ So $\log_k(x) = s = rt = \log_b(x)\log_k(b)$, from which you get the equality you want by dividing through by $\log_k(b)$.
For the second equality, $\log_b(x) = \log_b(k)\log_k(x)$, let $r=\log_b(x)$, $s=\log_b(k)$, and $t=\log_k(x)$. Then $b^r = x$, $b^s = k$, and $k^t=x$.
Hence we have: $$b^r = x = k^t = (b^s)^t = b^{st},$$ which means we must have $\log_b(x) = r=st=\log_b(k)\log_k(x)$, as desired.