Logarithms

Hong Kong

Stage 4 - Stage 5

Lesson

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`×10`k` for any integer $k$`k`.

For example, $\log_{10}\left(1.4\right)=0.14613...$`l``o``g`10(1.4)=0.14613... and $\log_{10}\left(14\right)=1.14613...$`l``o``g`10(14)=1.14613... and $\log_{10}\left(14000\right)=4.14613...$`l``o``g`10(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.

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`=`l``o``g``p``a`. We wish to express $y$`y` as a logarithm in base $q$`q`.

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

From the working rules of logarithms this simplifies to $\log_qa=y\log_qp$`l``o``g``q``a`=`y``l``o``g``q``p` and thus $y=\frac{\log_qa}{\log_qp}$`y`=`l``o``g``q``a``l``o``g``q``p`.

Look carefully at this result. It is saying that:

$\log_pa=\frac{\log_qa}{\log_qp}$`l``o``g``p``a`=`l``o``g``q``a``l``o``g``q``p`

For example, $\log_58=\frac{\log_{10}8}{\log_{10}5}$`l``o``g`58=`l``o``g`108`l``o``g`105, and so if we knew the common logs of $8$8 and $5$5, we could determine $\log_58$`l``o``g`58. Thus, since $\log_{10}8=0.90309$`l``o``g`108=0.90309 to five decimal places, and $\log_{10}5=0.69897$`l``o``g`105=0.69897, then $\log_58=\frac{0.90309}{0.69897}=1.29203$`l``o``g`58=0.903090.69897=1.29203.

Note that $\log_58$`l``o``g`58 could also be expressed as $\frac{\log_2\left(8\right)}{\log_2\left(5\right)}$`l``o``g`2(8)`l``o``g`2(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, $\log_b10=\frac{\log_{10}10}{\log_{10}b}=\frac{1}{\log_{10}b}$`l``o``g``b`10=`l``o``g`1010`l``o``g`10`b`=1`l``o``g`10`b` and so $\log_b10\times\log_{10}b=1$`l``o``g``b`10×`l``o``g`10`b`=1, which is an interesting result. We can generalise this to show that $\log_ba\times\log_ab=1$`l``o``g``b``a`×`l``o``g``a``b`=1 ,so that $\log_ba$`l``o``g``b``a` and $\log_ab$`l``o``g``a``b` are mutual inverses.

Change of Base Rule!

$\log_ab=\frac{\log_cb}{\log_ca}=\frac{1}{\log_ba}$`l``o``g``a``b`=`l``o``g``c``b``l``o``g``c``a`=1`l``o``g``b``a`

Rewrite $\log_416$`l``o``g`416 in terms of base $10$10 logarithms.

Rewrite $\log_320$`l``o``g`320 in terms of base $4$4 logarithms.

Rewrite $\log_3\sqrt{5}$`l``o``g`3√5 in terms of base $10$10 logarithms.