NZ Level 7 (NZC) Level 2 (NCEA)
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Using a Calculator for Logarithm Calculations
Lesson

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<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.

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, $\log_{10}52543=4.720515...$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=103 and $0.01=10^{-2}$0.01=102, 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 $\log_2\sqrt{3}=\frac{1}{2}\log_23=\frac{1}{2}\left(\frac{\log_{10}3}{\log_{10}2}\right)=0.7924812504$log23=12log23=12(log103log102)=0.7924812504.

Thus $10^{0.7924812504}=6.2012787052$100.7924812504=6.2012787052 and we can then write that $\log_2\sqrt{3}=\log_{10}\left(6.2012787052\right)$log23=log10(6.2012787052).

 

Worked Examples

Question 1

Consider the expression $\log8892$log8892.

  1. Within what range is the value of $\log8892$log8892?

    between $0$0 and $1$1

    A

    between $4$4 and $5$5

    B

    between $2$2 and $3$3

    C

    between $3$3 and $4$4

    D

    between $1$1 and $2$2

    E

    between $0$0 and $1$1

    A

    between $4$4 and $5$5

    B

    between $2$2 and $3$3

    C

    between $3$3 and $4$4

    D

    between $1$1 and $2$2

    E
  2. Find the value of $\log8892$log8892 correct to four decimal places.

Question 2

Find the value of $\log29^4$log294 correct to four decimal places.

Question 3

Consider the following.

  1. Rewrite $\log_821$log821 in terms of base-$10$10 logarithms.

  2. Hence evaluate $\log_821$log821 to two decimal places.

Outcomes

M7-6

Manipulate rational, exponential, and logarithmic algebraic expressions

91261

Apply algebraic methods in solving problems

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