Logarithms

(What's a Logarithm? How to Calculate Logarithms by Hand)


Logarithm is another word for exponent or power.

When we write: \[2^3 = 8\]

  • \(2\) is the base
  • \(3\) is the exponent and it is also called "the logarithm of \(8\), in base \(2\)". In other words:

    The logarithm is the power to which we have to raise the base (in this case \(2\)) in order to get the desired number (in this case \(8\)).

We can write that last statement in logarithmic form as follows: \[log_2(8)=3\] Read "log, base \(2\), of \(8\) equals to \(3\)", which literally means "the power to which we have to raise the base, \(2\), to get \(8\) is \(3\)".

The two statements: \[log_2(8)=3 \quad \text{and} \quad 2^3 = 8\] are equivalent statements.

The first is written in logarithmic form and the second in exponential form.

It is important to be comfortable writing expressions in either of these two forms.
This is further illustrated here:

More generally we write: \[log_b(a) = c \quad \text{when} \quad b^c = a\]

This is best explained in the following tutorial, watch it now before we learn ho to calculate logarithms.

Tutorial 1: What's a Logarithm?

How to calculate logarithms

Given a logarithm \(log_b(a)\), such as \(log_4(16)\), we'll often be required to evaulate (calculate) its value by hand.

To do this we start by writing the logarithmic equation: \[log_b(a)=x\] in its exponential form: \[b^x=a\]. Once this is done we solve the exponential equation for \(x\).

For \(log_4(16)\) we would start by defining the logarithmic equation \(log_4(16) = x\) and then write it in exponential form \(4^x = 16\) and solve for \(x\), which would be: \[4^x = 4^2\] And so: \[x = 2\] Which allows us to state \(log_4(16)=2\).

The method for calculating logarithms by hand is best explained in tutorial 2 below.

Tutorial 2: Calculating Logarithms by Hand

The following tutorial illustrates the method for calculating logarithms by hand.

Exercise

Calculate each of the following without a calculator:

  1. \(log_2(8)\)
  2. \(log_3(9)\)
  3. \(log_5(25)\)
  4. \(log_{10}(10000)\)
  5. \(log_4(64)\)
  6. \(log_3(81)\)
  7. \(log_2(32)\)

Answers Without Working

  1. \(log_2(8) = 3\)
  2. \(log_3(9) = 2\)
  3. \(log_5(25) = 2\)
  4. \(log_{10}(10000) = 4\)
  5. \(log_4(64) = 3\)
  6. \(log_3(81) = 4\)
  7. \(log_2(32) = 5\)


Logarithms of Decimals & Fractions

Now that we know how to calculate "basic" logarithms, we learn how to calculate logarithms of decimals and fractions.

For instance, we need to know how to calculate \(log_4(0.25)\) or \(log_5\begin{pmatrix}\frac{1}{25}\end{pmatrix}\).

Logarithms of Fractions & Decimals


To calculate logarithms such as \(log_4(0.25)\) or \(log_5\begin{pmatrix}\frac{1}{25}\end{pmatrix}\), we'll need to be comfortable write decimals and fractions as negative exponents. We start by reminding ourselves of these here.

Reminder: Negative Powers


For any non-zero number \(a\), the following will always be true: \[\frac{1}{a^n} = a^{-n}\] For instance, \(\frac{1}{3^2} = 3^{-2}\).

Using the result, above, we can calculate the logarithms, such as \(log_4(0.25)\) or \(log_5\begin{pmatrix}\frac{1}{25}\end{pmatrix}\), by writing the number inside the logarithm as a power of the base.

How to use negative powers with logarithms


Note: when we're given a decimal, like \(log_4(0.25)\), it will often be more convenient to start by writing the decimal as a fraction: \(log_4(0.25) = log_4\begin{pmatrix}\frac{1}{4}\end{pmatrix}\).

Example

Exercise

Calculate each of the following without a calculator:

  1. \(log_{10}(0.1)\)
  2. \(log_2(0.5)\)
  3. \(log_2(0.25)\)
  4. \(log_4(0.25)\)
  5. \(log_5(0.2)\)
  6. \(log_{10}(0.001)\)

Answers Without Working

  1. \(log_2(8) = 3\)
  2. \(log_3(9) = 2\)
  3. \(log_5(25) = 2\)
  4. \(log_{10}(10000) = 4\)
  5. \(log_4(64) = 3\)
  6. \(log_3(81) = 4\)
  7. \(log_2(32) = 5\)

Tutorial 3: Logarithms of Decimals & Fractions

Exercise

Calculate each of the following without a calculator:

  1. \(log_{10}(0.1)\)
  2. \(log_2(0.5)\)
  3. \(log_2(0.25)\)
  4. \(log_4(0.25)\)
  5. \(log_2(0.125)\)
  6. \(log_5(0.2)\)
  7. \(log_{10}(0.001)\)

Answers Without Working

We find the following:

  1. \(log_{10}(0.1) = -1\)
  2. \(log_2(0.5) = -1\)
  3. \(log_2(0.25) = -2\)
  4. \(log_4(0.25) = -1\)
  5. \(log_2(0.125) = -3\)
  6. \(log_5(0.2) = -1\)
  7. \(log_{10}(0.001) = -3\)


Tutorial 4: when the base is greater than the number & the result is a fraction

Exercise

Calculate each of the following without a calculator:

  1. \(log_4(2)\)
  2. \(log_8(2)\)
  3. \(log_{81}(3)\)
  4. \(log_{81}(9)\)
  5. \(log_{64}(4)\)
  6. \(log_{49}(7)\)

Answers Without Working

We find the following:

  1. \(log_4(2) = \frac{1}{2}\)
  2. \(log_8(2) = \frac{1}{3}\)
  3. \(log_{81}(3) = \frac{1}{4}\)
  4. \(log_{81}(9) = \frac{1}{2}\)
  5. \(log_{64}(4) = \frac{1}{3}\)
  6. \(log_{49}(7) = \frac{1}{2}\)


More complicated cases

At times it is less obvious to express the inpot value of a logarithm as a power of the base. For instance, if we were asked to evaluate: \[log_9(27)\] The power \(x\) to which we need to raise the base \(9\) for it to equal to \(27\) isn't neccessarily obvious.

Nevertheless we can see that both the base \(9\) and the input value \(27\) can be expressed as powers of \(3\) and that's what will allow us to evalate this logarithm.

This is further explained in the tutorial below, spend a few minutes to watch it.

Tutorial 5: when the result is an improper fraction

Exercise

Answers Without Working