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Logarithms


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Numeracy Basics

Logarithms

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This topic will cover how to:

  • convert between exponents and logarithms;
  • evaluate logarithms manually;
  • evaluate logarithms with base 10 or base e on a scientific calculator; and
  • apply five logarithm laws to rearrange and simplify logarithms.

Logarithms

Logarithms are the 'opposite' of exponents, in the same way as addition and subtraction are opposites and multiplication and division are opposites; that is, the two operations undo each other.

Hence if you have an equation that sets a positive number a equal to an exponent with positive base band some integer power c it can be rearranged to an equation that sets c equal to the logarithm of a with base b :

a = bc, then c = logb(a)

This relationship means that logarithms can be used to solve for an unknown exponent in an exponential equation- which is one of the reasons you may need to use them.

While you simply need to be able to remember this relationship at this stage, by the completion of this topic you should be able to understand how to convert from one equation to the other using the logarithm laws.

a =

bc

=then c

=logb(a)

Evaluating Logarithms Manually

The logarithm of a number with a certain base tells you how many times you need to multiply the base by itself in order to obtain the number in question. For example, evaluating logba tells you how many times you need to multiply b by itself in order to obtain a. So just how do you evaluate this?

For most bases, evaluating the logarithm simply requires you to perform repeated multiplication of the base until the correct answer is obtained. For example, to evaluate log216 you would determine how many times 2 needs to be multiplied by itself in order to equal 16. Repeated multiplication gives a solution of 4, since:

2 x 2 x 2 x 2 = 2 4 = 16

Therefore we have:

log2(16) = 4

This tells us that the logarithm of 16 with base 2 is 4, and you should observe that the relationship between the logarithm and exponent are as described previously.

2 x 2 x 2 x 2 =

2 4

= 16

log2(16) = 4

Examples: Evaluating Logarithms Manually

Evaluate the following logarithms:

1. log5(125)

5 x 5 x 5 = 53 = 125

So log5(125) = 3

2. log4(1024)

4 x 4 x 4 x 4 x 4 = 45 = 1024

So log4(1024) = 5

3. Rearrange log3(x) = 5 to give an equation for x in terms of an exponent, and hence solve for x

Rearranging gives x = 35

Hence x = 243

5 x 5 x 5 = 53 = 125

So log5(125) = 3

4 x 4 x 4 x 4 x 4 = 45 = 1024

So log4(1024) = 5

Rearranging gives x = 35

Hence x = 243

Activity 1: Practice Questions

Evaluating Logarithms with Base 10

There are two bases that are used most commonly with logarithms, and there are buttons on a scientific calculator for evaluating these. The first of these bases is 10, and in fact if you see a logarithm written without any base at all you can usually assume the base to be 10. For example:

log(100 000) = log10(100 000)

The button for evaluating a log with base 10 on a scientific calculator is the 'log' button, so to evaluate the logarithm above you would enter 100 000 and then press 'log', giving a result of 5:

log(100 000) = 5

log(100 000)

= log10(100 000)

log(100 000)

= 5

Evaluating Logarithms with Base e

The other commonly used base for logarithms is a special number known as Euler's number, which is denoted by the letter e. e is an irrational number that is very important in mathematics, economics and finance, and its first 20 digits are:

2.7182818284590452353...

A logarithm with base e is referred to as a natural logarithm, and is generally denoted by ln. For example:

ln(7.389) = loge(7.389)

The button for evaluating a log with base e on a scientific calculator is the 'ln' button, so to evaluate the logarithm above you would enter 7.389 and then press 'ln', giving a result of approximately 2:

ln(7.389)= ≈ 2

ln(7.389)

= loge(7.389)

ln(7.389)

≈ 2

Examples: Evaluating Logarithms
with Base 10 or e

Evaluate the following logarithms, rounding to two decimal places as necessary:

1. log(500)

Evaluating using a calculator gives 2.70

2. ln(200 000)

Evaluating using a calculator gives 12.21

3. Rearrange 400 = ex to give an equation for x in terms of a natural logarithm, and hence solve for x

Rearranging gives x = loge(400) = ln(400)

Hence x = 5.99

Evaluating using a calculator gives 2.70

Evaluating using a calculator gives 12.21

Rearranging gives x = loge(400) = ln(400)

Hence x = 5.99

Activity 2: Practice Questions

Logarithm Laws 1 and 2

As with exponents, there are some 'laws' you need to abide by when working with logarithms. We will cover the five most important ones here.

When detailing these laws, we will use the variables a, b and c to represent any positive numbers.

Logarithm Laws 1 and 2 are the simplest two logarithm laws, and are as follows:

Law 1 :

logb(b) = 1

For example, log10(10) = 1

Law 2 :

logb(1) = 0

For example, log10(1) = 0

logb(b) = 1

For example, log10(10) = 1

logb(1) = 0

For example, log10(1) = 0

Logarithm Laws 3 and 4

Logarithm Laws 3 and 4 cover how to deal with logarithms of products and divisions respectively, and are as follows:

For example, log2(4 x 8) = log2(4) + log2(8)

= 2 + 3

= 5

= log2(32)

Law 4 :

logb(a/c)= logb(a) - logb(c)

For example, log2(64/4) = log2(64) - log2(4)

= 6 - 2

= 4

= log2(16)

Law 3 :
logb(ac)= logb(a) + logb(c)

For example, log2(4 x 8) =

log2(4) + log2(8)

= 2 + 3

= 5

= log2(32)

Law 4 :
logb(a/c)= logb(a) - logb(c)

For example, log2(64/4) = log2(64) - log2(4)

= 6 - 2

= 4

= log2(16)

Logarithm Law 5

Logarithm Law 5 covers logarithms of exponents, and is as follows:

Law 5 :

logb(ac)= clogb(a)

For example, log2(82)= 2 log2(8)

= 2(3)

= 6

= log2(64)

Law 5 :

logb(ac)= clogb(a)

For example, log2(82)=

2 log2(8)

= 2(3)

= 6

= log2(64)

Examples: Logarithm Laws

1. Evaluate log15(15) + log3(3)

log15(15) + log3(3)= 1 + 1

= 2

2. Express log(x) + log(y) - 2log(z) as a single logarithm

log(x) + log(y) - log(z) = log(xy) - 2log(z)

= log(xy) - log(z 2)

= log(xy/z^2)

3. Evaluate log64 + log654 + log7(1) =

log64 + log654 + log7(1) = log64 + log654 + 0

= log64 + log654

= log6(4 x 54)

= log6(216)

= 3

log15(15) + log3(3)= 1 + 1

= 2

log(x) + log(y) - log(z) = log(xy) - 2log(z)

= log(xy) - log(z 2)

= log(xy/z^2)

log64 + log654 + log7(1) = log64 + log654 + 0

= log64 + log654

= log6(4 x 54)

= log6(216)

= 3

Activity 3: Practice Questions

Exponentials and Logarithms

Now that we have covered the logarithm laws, we can look at how to derive the relationship between exponentials and logarithms. This is as follows:

a =b c

logb(a) = logb(b c)

logb(a) = clogb(b) (by Logarithm Law 5)

logb(a) = c(1) (by Logarithm Law 1)

logb(a) = c

c = logb(a)

a =b c

logb(a) = logb(b c)

logb(a) = clogb(b) (by Logarithm Law 5)

logb(a) = c(1) (by Logarithm Law 1)

logb(a) = c

c = logb(a)

End of Topic

Congratulations, you have completed this topic.

You should now have a better understanding of Logarithms.

 


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