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Skills for Learning

Maths & Stats - Powers and Logarithms

Learning Outcomes

If you work through this section you should be able to:

  • Explain the term 'logarithm'.
  • Understand the rules of logarithms and use them to simplify expressions.
  • Show how to change the base.
  • Solve exponential and logarithmic equations.
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A power or exponent of a number states how many times to multiply the number by itself. The power is called the logarithm of a number to the base.

For example, 25 = 32, then 5 = log2⁡ 32, which is spoken as 5 is the logarithm of 32 to the base 2.

Examples

Exponential form

Logarithmic form

34 = 81

4 = log3 81

63 = 216

3 = log6 216

105 = 100000

5 = log10 100000

Because logarithms are powers, the rules of logarithms closely follow the rules of indices. The rules of logarithms are given as below:

loga(xy)=logax  +  logay

loga(xy)=logax - logay

loga(x)n=nlogax

logax=logbxlogba

logaa= 1

loga1= 0

Examples

1. log25 + log23 = log2(5x3) = log215

2. log310 - log32 = log3(102) = log35

3. log1053 = 3log105

4. log106 + log1014 - log107 + 3log102 = log106 + log1014 - log107 + log1023

   = log106 + log1014 - log107 + log108

   = log10(6×14×8/7)

   = log1096

5. log48 = log28log24

    = 32

    = 1.5

When a logarithm is written without a base, it usually means that the base is 10. It is called common logarithm. For example, log⁡100 = log10⁡100 = 2.

Another base that is often used is e (Euler's Number) which is about 2.71828. Logarithms to base e are called natural logarithms, which are denoted by loge or ln. For example, ln(7.389) = loge(7.389) ≈ 2.

Negative logarithms state how many to divide by the number.

For example, 2-4 = 124 = 0.0625, therefore log2 0.0625 = −4.

We will now look at how to use logarithms to solve logarithmic equations.

Example 1

Solve the following for x:

log2(3x1)=3

Solution

log2(3x1)=3 so that

3x-1= 23 hence

3x-1=8 that is

3x=8+1 hence

3x=9 hence

x=3

Example 2

Solve the following for x:

2loga2x-loga(x-1) = loga(4x-1)

Solution

2loga2x-loga(x-1) = loga(4x-1) so that

loga2x2-logax-1 = loga4x-1 that is

loga4x2 - loga(x-1) = loga(4x-1) that is

loga4x2(x-1) = loga(4x-1) so that

4x2(x-1) =4x-1 that is

4x2 = (x-1)×(4x-1) that is

4x2 = 4x2-5x+1 that is

5x = 1 hence

x = 0.2

Example 3

Find y in terms of x:

logax+1 + 2logay = 3logay-loga3x + logax

Solution

logax+1 + 2logay = 3logay-loga3x + logax so that

logax+1+loga3x-logax = 3logay-2logay that is

logax+1 + loga3x-logax = logay so that

logax+1×3xx = logay so that

x+1×3xx = y that is

y = x+1×3 hence

y = 3x+3

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