Introduction to Logarithms (Logs)

Logarithms or logs are another way of expressing indices and is a bit like the inverse of dealing with powers. Logs were used to perform difficult calculations before there were calculators.

To understand how logs work we could just look at a simple to understand example:

10^{2}=100 \rightarrow Log_{10} (100)=2

We know that 10^{2}=100, so if we wanted to know the value of the indices that would make 10 into 100 (how many times 10 multiplied by itself gives us 100 i.e. 2) then we could use the above to work it out. We would say ‘Log to the base 10 of 100 is 2, or other words, 10 squared is 100‘.

If we look at the original expression 10^{2}=100 and transform that into Log language (how you would input it into your calculator) we’d say ‘LOG TO THE BASE of the number that has a power, OF the answer that we’d get from the indices IS the power’.

Logs can’t be done with negative numbers and always have to be done with positive numbers. If we’ve got a value for a negative log then that is for a value for an indices that is less than 1 i.e. 10^{-1}=0.1 \rightarrow Log_{10} (0.1)=-1 because when we deal with indices, if we’ve got a negative power that means we’ve got a fraction.


Log in base 10 (Decimal)

Logs work for all powers of 10 and not just integers:

10^{y}=x

Log_{10} (x)=y

Therefore:

10^{0.477}=3

Log_{10} (3)=0.477

y=x (Green) vs y=Log_{10} (x) (Red)

If we look at a comparison (shown in the picture above) of y=x & y=Log_{10} (x) we’ll see that the curve of the Red line (y=Log_{10} (x)) is very steep at the begining but falls off after (1,1) (in this instance) below the Green line (y=x).


ln – base e

Natural (Napierian) logs work in exactly the same way as base 10 logs but because we are dealing with natural number e we can use ln:

e^{y}=x

ln (x)=y

Just like base 10 logs  \begin{cases}ln(1)=0 \\ ln(e)=1 \end{cases}


Rules of logs

There are three rules or laws of logs that can be used in all bases:

ln (x) + ln (y)= ln (x \cdot y)

$ln (x) - ln (y)= ln (x \div y)$

a \cdot ln (x) = ln (x^{a})


Rule 1 Proof

Let ln (x) = a, ln (y)= b \rightarrow e^a = x, e^{b}= y

e^a \cdot e^b = e^{a+b} = x \cdot y

a + b = ln(x \cdot y)

ln(x) + ln(y) = ln(x \cdot y) // Substitute in a & b with ln(x) and ln(y)


Rule 2 Proof

Let ln (x) = a, ln (y)= b \rightarrow e^a = x, e^{b}= y

e^a \div e^b = e^{a-b} = x \div y

a + b = ln(x \div y)

ln(x) - ln(y) = ln(x \div y) // Substitute in a & b with ln(x) and ln(y)


Rule 3 Proof

ln(x^a) = ln(x \cdot x \cdot  x \cdot  x \cdot x) // a times

ln(x^a) = ln(x) + ln(x) + ln(x) + ln(x) + ln(x) // a times

a \cdot ln (x) = ln (x^{a})


Cahnging bases

Logs can be done in any base but calculators only have the option for base 10 or base e so if we need to calculate a different base then we would need to convert the base we want into one fo these two first.

As an example let’s evaluate lg_2 (3) = x.

Firstly we would rearrange to get 2^x = 3 and then take logs (base 10 or e) x ln(2) = ln (3). Then simply rearrange for x to get x = ln(3) \div ln(2) = 1.585.

Leave a Comment

Your email address will not be published. Required fields are marked *