NZ Level 7 (NZC) Level 2 (NCEA)
Change of Base
Lesson

## Common Logs

Common logarithms, first devised by Henry Briggs in 1620, were base $10$10 logarithms. They had space saving advantages because the log of a number $N$N would have the same decimal part of the log of $N\times10^k$N×10k for any integer $k$k.

For example, $\log_{10}\left(1.4\right)=0.14613...$log10(1.4)=0.14613... and $\log_{10}\left(14\right)=1.14613...$log10(14)=1.14613... and $\log_{10}\left(14000\right)=4.14613...$log10(14000)=4.14613...etc, all have the same decimal component of $0.14613...$0.14613...

The answers to logarithm evaluations used to be stored in a logarithm table, they looked a little like this

Today, modern computing devices have made common logarithm tables redundant, and many devices can evaluate a logarithm to any desired base.

## Changing the base

We often encounter occasions where we need to take a logarithm given in one base and express it as a logarithm in another base. A change of base formula has been developed to do just that.

Suppose we think of a number $y$y expressed as $y=\log_pa$y=logpa. We wish to express $y$y as a logarithm in base $q$q

Since $y=\log_pa$y=logpa, from the definition of a logarithm, this means that $a=p^y$a=py. If we now take logarithms to base $q$q on this last equation, we have that $\log_qa=\log_q\left(p^y\right)$logqa=logq(py).

From the working rules of logarithms this simplifies to $\log_qa=y\log_qp$logqa=ylogqp and thus $y=\frac{\log_qa}{\log_qp}$y=logqalogqp

Look carefully at this result. It is saying that:

$\log_pa=\frac{\log_qa}{\log_qp}$logpa=logqalogqp

For example, $\log_58=\frac{\log_{10}8}{\log_{10}5}$log58=log108log105, and so if we knew the common logs of $8$8 and $5$5, we could determine  $\log_58$log58. Thus, since $\log_{10}8=0.90309$log108=0.90309 to five decimal places, and $\log_{10}5=0.69897$log105=0.69897, then $\log_58=\frac{0.90309}{0.69897}=1.29203$log58=0.903090.69897=1.29203

Note that $\log_58$log58 could also be expressed as $\frac{\log_2\left(8\right)}{\log_2\left(5\right)}$log2(8)log2(5) which would give the same answer $1.29203$1.29203. Any base can be used, now we have the above relationship.

As another example, $\log_b10=\frac{\log_{10}10}{\log_{10}b}=\frac{1}{\log_{10}b}$logb10=log1010log10b=1log10b and so $\log_b10\times\log_{10}b=1$logb10×log10b=1, which is an interesting result. We can generalise this to show that $\log_ba\times\log_ab=1$logba×logab=1 ,so that $\log_ba$logba and $\log_ab$logab are mutual inverses.

Change of Base Rule!

$\log_ab=\frac{\log_cb}{\log_ca}=\frac{1}{\log_ba}$logab=logcblogca=1logba

#### Worked Examples

##### QUESTION 1

Rewrite $\log_416$log416 in terms of base $10$10 logarithms.

##### QUESTION 2

Rewrite $\log_320$log320 in terms of base $4$4 logarithms.

##### QUESTION 3

Rewrite $\log_3\sqrt{5}$log35 in terms of base $10$10 logarithms.

### Outcomes

#### M7-6

Manipulate rational, exponential, and logarithmic algebraic expressions

#### 91261

Apply algebraic methods in solving problems