Use a calculator to evaluate logarithms

A little history first

Henry Briggs, in 1624, evaluated and published the base $10$10 logarithms of thirty thousand numbers, all worked  out by hand to $14$14 decimal places. It was a triumph of mathematical calculation. 

So popular did Briggs' tables become that, still to this day, the mathematical expression $\log a$loga, where the base is not mentioned, is by convention a base $10$10 logarithm. 

Even though modern calculators provide us with logarithm values, there are at least two concepts to come out of that work that are extremely useful to us today.  

Concept 1 

Firstly, in working out a number like $\log2$log2 base $10$10, Briggs realised that a number like $2^9=512$29=512 is between $10^2=100$102=100 and $10^3=1000$103=1000. 

To derive an estimate of $\log2$log2, he wrote:  

$10^2<2^9<10^3$102<29<103 

Then, applying logarithms base $10$10, he wrote:

$\log_{10}\left(10^2\right)<\log_{10}\left(2^9\right)<\log_{10}\left(10^3\right)$log10​(102)<log10​(29)<log10​(103)

and this simplifies to:

$2<9\log_{10}2<3$2<9log10​2<3

He then divided this last inequality by $9$9 to get his first estimate of $\log2$log2 - a number that was larger than $\frac{2}{9}=0.222...$29​=0.222... but smaller than $\frac{1}{3}=0.333...$13​=0.333....

By using larger powers of $2$2 the interval where $\log2$log2 lies becomes smaller and smaller. To $6$6 figures we know that $\log2=0.301030$log2=0.301030.

 

Considering another example:

$100000$100000	$<$<	$531441<1000000$531441<1000000
$10^5$105	$<$<	$531441<10^6$531441<106
$5$5	$<$<	$\log531441<6$log531441<6
$5$5	$<$<	$\log3^{12}<6$log312<6
$5$5	$<$<	$12\log3<6$12log3<6
$\frac{5}{12}$512​	$<$<	$\log3<\frac{1}{2}$log3<12​

So $\log3$log3 is a number larger than $\frac{5}{12}=0.41666...$512​=0.41666... but smaller than $\frac{1}{2}=0.5$12​=0.5. We know that, to $6$6 decimal places, $\log3=0.477121$log3=0.477121.

Concept 2

The second concept we learn from Briggs is that if we know that $\log2=0.301030$log2=0.301030 and $\log3=0.477121$log3=0.477121, then we also know that:

$\log6$log6	$=$=	$\log\left(2\times3\right)$log(2×3)
	$=$=	$\log2+\log3$log2+log3
	$=$=	$0.301030+0.477121$0.301030+0.477121
	$=$=	$0.778151$0.778151

In other words, Briggs realised that if he evaluated base $10$10 logs of prime numbers  as accurately as possible, he also could obtain logs of composite numbers containing those primes as factors. For example $\log2592=\log\left(2^5\times3^4\right)=5\log2+4\log3$log2592=log(25×34)=5log2+4log3.  

 

Estimates of logs

Take a number like $52543$52543. Since $10000=10^4$10000=104 and $100000=10^5$100000=105, then $4<\log52543<5$4<log52543<5. Thus we know that $\log52543$log52543 is "$4$4 point something". In fact, $\log_{10}52543=4.720515...$log10​52543=4.720515.... The whole number $4$4 was referred to as the characteristic of the logarithm and the decimal part $0.720515...$0.720515... became known as the mantissa of the logarithm. 

Take a number like $0.00734$0.00734. Since $0.001=10^{-3}$0.001=10−3 and $0.01=10^{-2}$0.01=10−2, then $-3<\log0.00734<-2$−3<log0.00734<−2. Thus we know that $\log0.00734$log0.00734 is "minus 2 point something" and knowing this allows us to check the sense of any value we access from a calculator.  

 

Other bases

We can change a logarithm of any base to a base $10$10 logarithm by using the change of base rule and a modern scientific calculator.

 

Example 1

For example we might ask what $\log_3\left(28\right)$log3​(28) is expressed as a base $10$10 logarithm. 

Working to $10$10 decimal places we have $\log_3\left(28\right)=\frac{\log_{10}\left(28\right)}{\log_{10}\left(3\right)}=3.0331032563$log3​(28)=log10​(28)log10​(3)​=3.0331032563.

Now $10^{3.0331032563}=1079.2032792454$103.0331032563=1079.2032792454 and so we can say that $\log_3\left(28\right)=\log_{10}\left(1079.2032792454\right)$log3​(28)=log10​(1079.2032792454).   

 

Example 2

As another example $\log_2\sqrt{3}=\frac{1}{2}\log_23=\frac{1}{2}\left(\frac{\log_{10}3}{\log_{10}2}\right)=0.7924812504$log2​√3=12​log2​3=12​(log10​3log10​2​)=0.7924812504.

Thus $10^{0.7924812504}=6.2012787052$100.7924812504=6.2012787052 and we can then write that $\log_2\sqrt{3}=\log_{10}\left(6.2012787052\right)$log2​√3=log10​(6.2012787052).

 

Worked Examples

Question 1

Question 2

Question 3