# Logarithms

Because 'log' kya sochenge...

# Properties of Logarithms

## 1. Def:$\log_ba$ is read as "log a to the base b", or in short as "log a base b" where a is the argument and b is the base

$(base)^{(power)}=(argument) \;\;\;\; =\;\;\;\; \log_{(base)}(argument)$

## 2. $\log_{b}a$ is defined iff $a > 0, b > 0, b \neq 1$​

iff means 'if and only if' in standard mathematics

Note that $b>0, b \neq 1$cannot be written as $b>1$, as $b$can still be between 1 and 0

Why can't base be 1? In essence, Log finds the power given the value and base. If base is 1, value cannot be anything other than 1, as $1^{anything}$is still 1.

## 3. $\log_aa=1$​

Proof: Anything raised to 1 is the same thing.

## 4. $\log_a1 =0$​

Proof: If any base raised to some power is 1, then the power must be 0

## 5. $\log_ax+\log_ay=\log_a(xy)$​

Proof:

$\log_ax=m \longrightarrow a^m = x \newline \log_ay=n \longrightarrow a^n = y \newline$

Let's multiply the equations

$a^m\times a^n=xy$

We can do this because we're still multiplying both sides by the same value (LHS by $a^n$and RHS by $y$), as $a^n$is equal to $y$, i.e. the same as $y$

$a^{m+n}=xy \longrightarrow \log_{(a)}(xy)=m+n \newline \therefore \log_a(xy)=\log_ax+\log_ay$

## 6. $\log_ax - \log_ay = \log_a(\frac{x}{y})$​

Proof:

$\log_ax=m \longrightarrow a^m=x \newline \log_ay=n \longrightarrow a^n=y$

Divide the equations

$\frac{a^m}{a^n} = \frac{x}{y} \newline$

Which is

$a^{(m-n)} = \frac{x}{y} \longrightarrow \log_a(\frac{x}{y}) = m-n \newline \therefore \log_a(\frac{x}{y}) = \log_ax - \log_ay$

## 7. $a^{\log_ab} = b$​

Proof: Let the answer be x

$a^{\log_ab} = x$

Can be written as

$\log_ax=\log_ab$

...using property 1. Since bases are the same powers can be equated

$\therefore x=b$

## 8. $\log_a{b^k} = k \times \log_ab$

Proof:

Let $m=\log_ab$

$b^m=a$

...using property 7.

$(b^m)^k=(a)^k$

Which is

$b^{m\times k} = a^k$

Using property 1...

$\log_ab^{k}=m\times k$

As $m=\log_ab$

$\log_a{b^k} = k \times \log_ab$

## 9. $\frac{\log_ca}{\log_cb} = \log_ba$​

Proof:

Let $\log_ba=x$. Using property 7...

$b^x = a$

Take $\log_c$on both sides

$\log_c{b^x} = \log_ca$

Using property 8...

$x \times \log_cb=\log_ca$

Divide both sides by $\log_cb$

$x=\frac{\log_ca}{\log_cb}$

Since $x=\log_ba$

$\frac{\log_ca}{\log_cb}= \log_ba$

## 9.1 $\log_ba = \frac{1}{log_ab}, \text{if} \; c=a$​

Proof:

Using property 9,

$\frac{\log_ca}{\log_cb}=\log_ba$

Since in this case $c=a$

$\frac{\log_aa}{\log_ab}=\log_ba$

Using property 7,

$\frac{1}{\log_ab}=\log_ba$

## 10. $\log_{b^k}a = \frac{1}{k} \times \log_ba$​

Proof:

Unwrap using property 9...

$\log_{b^k}a=\frac{\log_ca}{\log_cb^k}$

Extract $k$

$= \frac{\log_ca}{k \times \log_cb}$

Split the fraction

$= \frac{1}{k} \times \frac{\log_ca}{\log_cb}$

Re-wrap using property 9...

$\log_{b^k}a=\frac{1}{k} \times \log_ba$

## 11. $\log_{b^n}{a^m}=\frac{m}{n} \times\log_ba$​

This property compiles properties 9 and 10.

## 12. $a^{\log_cb}=b^{\log_ca}$​

Proof: Advanced, I may add it later