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.
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<9log102<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.
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.
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, $\log_{10}52543=4.720515...$log1052543=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.
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.
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).
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=12log23=12(log103log102)=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).
Consider the expression $\log8892$log8892.
Within what range is the value of $\log8892$log8892?
between $0$0 and $1$1
between $4$4 and $5$5
between $2$2 and $3$3
between $3$3 and $4$4
between $1$1 and $2$2
between $0$0 and $1$1
between $4$4 and $5$5
between $2$2 and $3$3
between $3$3 and $4$4
between $1$1 and $2$2
Find the value of $\log8892$log8892 correct to four decimal places.
Find the value of $\log29^4$log294 correct to four decimal places.
Consider the following.
Rewrite $\log_821$log821 in terms of base-$10$10 logarithms.
Hence evaluate $\log_821$log821 to two decimal places.
Manipulate rational, exponential, and logarithmic algebraic expressions
Apply algebraic methods in solving problems