Science, Maths & Technology

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# 2 Three log rules

The three log rules can be deduced from the rules of indices that you learned in Week 1.

If you let loga x = m and loga y = n

This means that x = amand y = an

So xy = am × an = a(m + n) (from Week 1: Rule 1).

Applying the definition of a logarithm gives:

loga xy = loga a(m + n)

loga xy = m + n

loga xy = loga x + loga y

For example, loga 21 = loga 7 + loga 3, since 7 and 3 are both factors of 21.

Again, let logax = m and logay = n with x = am and y = an.

(from Week 1: Rule 2)

Applying the definition of a logarithm gives:

loga  = loga(a(m – n))

loga  = m – n

For example, loga 7 = loga 14 – loga 2.

Let logax = n with x = an

Raising each side to the power r gives

(from Week 1: Rule 3)

Applying the definition of a logarithm gives:

logaxr = loga(arn)

For example, log10 1000 = log10 103 = 3 log10 10 = 3 (because log10 10 = 1)

Use the rules in this next activity.

## Activity 3 Using the three log rules

1. Simplify  log 6 + log 3 – log 9
2. Write  4 log x – ½ log y + 3 log z  as a single logarithm
3. log 64 ÷ log 2 (Note it can be useful to consider if a number can be written in the 2n)
4. (log 27 − log 9) ÷ log 3

### Discussion

1. Using Rule 1: log (6) + log (3) = log (6 × 3)

Giving: log (18) – log (9)

Using Rule 2: log (18) – log (9) = log (2)

So, log (6) + log (3) = log (2)

2. Using Rule 3: 4 log x – ½ log y + 3 log z = log x4 – log y½ + log z3

Then using Rule 2:

The Rule 3:
3. 64 = 26

log (64) ÷ log (2) = log (26) ÷ log (2)

Using rule 3: log (26) = 6 log (2)

log (26) ÷ log (2) = 6 log (2) ÷ log (2)

= 6

4. Using Rule 2: log (27) – log (9) = log (27 ÷ 9)

In the next section you will learn about a special logarithm, called the natural logarithm.