Proof 19: Logarithm rules of algebra


A logarithm finds the exponent $a$ that when put on a base $b$ gives the number. In other words, it ‘takes out’ the exponent:


If no base is written, we usually by default mean base 10. And in the special case where the base is the natural number $e$, we usually rename the logarithm from $\log$ to $\ln$ (as a ‘natural logarithm’):

\ln(a)=\log_{e}(a) \quad_\blacksquare $$

The exponent that gives the base itself is of course just 1 ($a^1$ is the number $a$ itself):


Regarding powers, we chose to define $a^0=1$. So, the logarithm to the number 1 will find the exponent 0, no matter the base since any base raised to 0 will give 1:



Also note that when the exponent is zero, it does not make the number become zero. In fact, no exponent can make the number zero! Also, regarding powers, we chose to define negative exponents as fractions: $a^{-b}=\frac 1{a^b}$. A negative exponent does not make the number negative – it makes it into a small but positive number, a fraction. In fact, no exponent can make the number negative either!

So, you cannot input a zero or a negative number into a logarithm; the logarithm will try to find the exponent that can create this zero or negative number, but no such exponent exists. It is thus a requirement for any $\log(a)$ that $a>0$.$\quad_\blacksquare$


100 is the same as ‘10 to the power of 2’, $100=10^2$. 90 can also be written as a power of 10, although not as pretty: $90=100^{1.9542}$. Any number can be rewritten as a power of 10: $a=10^c$.

This makes us able to show the logarithm rules that ‘multiplied inside means added outside’ and ‘divided inside means subtracted outside’ (we show base 10, but it would work for any base):

$$\log(a\cdot b)=\log(\underbrace{10^c}_a\cdot \underbrace{10^d}_b)=\log(10^{c+d})=c+d=\underbrace{\log(10^c)}_c+\underbrace{\log(10^d)}_d=\log(a)+\log(b)\quad_\blacksquare$$

$$\log\left(\frac ab\right)=\log\left(\frac{10^c}{10^d} \right )=\log\left(10^{c-d} \right )=c-d=\underbrace{\log(10^c)}_c-\underbrace{\log(10^d)}_d=\log(a)-\log(b) \quad_\blacksquare $$

Also, from these rules we can show that ‘powers inside mean multiplication outside’ and from that rule that ‘roots inside mean division outside’:

$$\log(a^b)=\log\underbrace{(a\cdot a \cdot a \cdots)}_{b \text{ many terms}} = \underbrace{\log(a)+\log(a)+\log(a)+\cdots}_{b \text{ many terms}}=b\cdot \log(a)\quad_\blacksquare$$

$$\log\left(\sqrt[b]a\right)= \log\left(a^{1/b}\right)=\frac 1b\cdot \log(a) \quad_\blacksquare$$