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## Rules of Logarithms

### Calculating Log values

Lets look at an example log.

$log_{10}100=2$

This tells us that $10^2=100$

The general form is $log_{base}answer=index\Rightarrow base^{index}=answer$

Examples

Calculate the following

$log_5125$

We need to find out what power we raise 5 to that gives the answer 125.

Since $125=5^3$ it is clear that $log_5125=3$

$log_3\frac{1}{9}$We need to find out what power we raise 3 to that gives the answer $\frac{1}{9}$

$\frac{1}{9}=9^{-1}=(3^2)^{-1}=3^{-2}$

So $log_3\frac{1}{9}=-2$

$log_100.001$We need to find out what power we raise 10 to that gives the answer $log_100.001$

$0.001=\frac{1}{1000}=\frac{1}{10^3}=10^{-3}$

So $log_100.001=-3$

$log_a1$Anything to the power of zero equals 1.

So it follows that any log of 1 0.

Therefore $log_a1=0$

$log_aa$Anything to the power of 1 equals itself.

So if follows that $log_aa=1$

$log_39+log_28$$log_39=2\because 3^2=9$$log_28=3\because 2^3=8$$\implies log_39+log_28 = 5$

$log_7{-49}$At first this looks quite easy, but there is no value you can raise a positive value to that will give a negative value.So $log_7{-49}$ is impossible.

Proof

Let $a^b=x$ and $a^c=y$

$\Rightarrow log_ax=b$ and $log_ay=c$

$xy=a^b\times a^c=a^{b+c}$

$\Rightarrow log_axy=b+c$

$\Rightarrow log_axy=log_ax+log_ay$

You can show that $log_ax-log_ay= log_a\frac{x}{y}$ in the same way

Examples

Simplify the following

$log_57+log_54$

These are straight forward logs with the same base so we can join them with multiplication so the answer is $log_528$

$log 6-log 3$(if a base is not defined then any base will do and most people will just use base 10)

We can assume they have a base, because a base is not mentioned.

So the answer is $log 2\because \frac{6}{3}=2$

$log_36+log_{10}9$These are straight forward logs, but they have different bases so we can not join them simply

### Multiple logs (Powers)

Consider $log x + log x + \dots +log x \text{ n times}$

$=nlog x$

Using the addition rule it also equals $log x^n$

So $nlog x=log x^n$

Examples

Simplify

$2log x+log 4$

$2log x=log x^2$

So using the addition rule $2log x+log 4=log 4x^2$

Solve $3log x-log 2=log 4$$3log x =log x^3$

So using the subtraction rule we get $log \frac{x^3}{2}=log 4\implies \frac{x^3}{2}=4\implies x^3=8\implies x=2$

### Solving equations with variable powers

We can solve equations with variable powers by:

1. isolating the term with the variable
2. taking logs of both sides
3. using the multiple logs rule to take the power out of the log
4. dividing by the log and solving

Examples

Solve $5^x=17$

the $x\text{ term}$ is already isolated

taking logs of both sides gives $log 5^x=log 17$

using the multiple logs rule gives $x log 5=log 17$

dividing gives $x=\frac{log 17}{log 5}=1.76\text{ to 2 d.p.}$

Solve $7^{n-1}+9=27$subtracting 9 from both sides gives $7^{n-1}=18$$\implies log 7^{n-1}=log 18$$\implies (n-1)log 7=log 18$

$\implies n-1=\frac{log 18}{log 7}$

$\implies n=\frac{log 18}{log 7}+1=2.49\text{ to 2 d.p.}$

Solve $5^{n-1}=4\times 7^{n-1}$$\implies \frac{5^{n-1}}{7^{n-1}}=4$

$\implies \left(\frac{5}{7}\right)^{n-1}=4$

Taking logs and using the power rule gives $(n-1)log\left(\frac{5}{7}\right)=log 4$

$\implies n = \frac{log 4}{log\left(\frac{5}{7}\right)}+1=-3.12\text{ to 2 d.p.}$

Questions