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Rules of Logarithms

Calculating Log values

Lets look at an example log.


This tells us that 10^2=100

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


Calculate the following


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}


So log_3\frac{1}{9}=-2

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


So log_100.001=-3

log_a1Anything to the power of zero equals 1.

So it follows that any log of 1 0.

Therefore log_a1=0

log_aaAnything to the power of 1 equals itself.

So if follows that log_aa=1

log_39+log_28log_39=2\because 3^2=9log_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.

Adding Logs


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


Simplify the following


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}9These 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



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 43log 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


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=27subtracting 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.}


Categories: Algebra
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