## Why the Change of Base Formula Works For Logarithms and Exponents

I find I remember math concepts so much more if I understand why and how things actually work.  That said, why does the change of base formula work?  Check out the proof in this Wikipedia article.

Here’s another description:

Say we have $y=log_a b$ and you want to change the base to c, for example, say you only have a log base c button on your calculator, such as log base 10.  Well, $y=log_a b$ means that $a^y= b$.  If we want this in base c, that means we want $c^z= b$ where $z=log_ c b$.  Now check this out…we know that $a^y= b$…so we can just say, lets make $c^v= a$ where $v=log_c a$.  Then that would mean that $(c^v)^y=b$ since $c^v= a$ and $a^y= b$.  Now, we know how to solve an exponent of an exponent:$(c^v)^y=c^{vy}$.

So, we want $c^{vy}$, where we know $v=log_c a$ and $y=log_a b$ and thus $vy= log_c a * log_a b$.  So, $c^z= b$ and $c^{vy}=b$, thus $log_c b=vy$.  Then since $vy= log_c a * log_a b$, $log_c b=log_c a * log_a b$, and $log_c b$/$log_c a=$ $log_a b$.

Finally! We’ve converted $y=log_a b$ to another base, the base of $c$!

It may have been easier to label the starting logarithm as $y=log_a x$. Then $y=log_a x=log_a b * log_b x$. And $log_b x=log_a x/log_a b$. This may be easier to visualize: say $g=log_a b$ and $h=log_b x$. That means that $a^g=b$ and $b^h=x$. Then $x=b^h=(a^g)^h$. Then $y=log_a x=log_a (a^g)^h= h * log_a (a^g)$ by the power rule of logarithms, and since $log_a (a^g)=g$, that means $y=log_a x=g * h$, which means that $y=log_a x=g * h=log_a b * log_b x$.

Why does the power rule of logarithms work? Say $a^g=b$ and $b^h=x$. Say g=3. So $a^g$ means $a^3$ which is (a)(a)(a)=b. Then $b^h=x$, and say h=2. So $b^h=b^2$ means (b)(b). Well, since (a)(a)(a)=b, (b)(b)=(a)(a)(a)(a)(a)(a), see the associative property of multiplication for more info. So, that demonstrates an example of how $y=log_a x=log_a b^h=h * log_a b$. Since h=2 and $log_a b=g=3$, that’s 2*3=6, as shown by the result (a)(a)(a)(a)(a)(a).