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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) ?

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$\frac1{\log_k(b)}=\frac{\log_k(k)}{\log_k(b)}=\log_b(k)$ – J. M. Dec 13 '10 at 3:28
up vote 7 down vote accepted

Divide the second equation by $\log_b(k)$ and shuffle the variables to see that these are the same thing. The reason they're true can be expressed in terms of properties of exponents. Start with the observation that $\log_b(x)=y$ means that $b^y=x$. Then use $b=a^{\log_a(b)}$ and properties of exponents to get $b^y=a^{y\log_a(b)}=x$. The last equation says that $\log_a(x)=y\log_a(b)$. Now substitute $\log_b(x)$ back in for $y$ to get the desired identity.

This fits into the general principle that each of the fundamental properties of logarithms can be seen as a translation of a property of their inverses, the exponential functions. For example, logarithms convert products to sums ($\log_b(xy)=\log_b(x)+\log_b(y)$) because exponential functions convert sums to products ($b^{x+y}=b^xb^y$). Here, the fact that $b^x=a^{x\log_a(b)}$ says that you can change bases of your exponential function by multiplying the input variable by the constant $\log_a(b)$. In other words, exponential functions with different bases are related by horizontal stretches or shrinks of the graphs. When expressed in terms of the inverse functions, input and output switch, and the horizontal stretch (or shrink) changes to a vertical stretch (or shrink). That is, instead of multiplying the input by $\log_a(b)$, you divide the output by $\log_a(b)$ to get $\log_b(x)=\frac{\log_a(x)}{\log_a(b)}$. More succinctly, $g(x)=f(cx)$ implies $g^{-1}(x)=\frac{1}{c}f^{-1}(x)$.

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After seeing Bill Dubuque's nice one line derivation using the power rule, I thought I'd mention that you can also use the second identity (as proved in my or Arturo's answer) to quickly derive the power rule: $$\log_b(a^x)=\log_b(a)\log_a(a^x)=x\log_b(a).$$ – Jonas Meyer Dec 13 '10 at 5:25

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 $k^t=s$. So 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.

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This is an excellent explanation. – Quixotic Dec 13 '10 at 14:17

Yes. Your second equation is $\log_b k\cdot \log_k x = \log_b x$. Dividing by $\log_b k$ we get $\log_k x=\frac{\log_b x}{\log_b k}$. This is the same as your first equation with different symbols (an interchange of $k$ and $b$).

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+1,Nice one-liner :) – Quixotic Dec 13 '10 at 14:08

HINT $\: $ The first equation is $\rm\ log_K\ $ of $\rm\ B^{log_B\ X}\ =\ X$

and the $\:$ second $\:$ equation is $\rm\ log_B\ $ of $\rm\ K^{log_K\ X}\ =\ X$

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