1. Introduction to Logarithms
By Dan Chicos. Last modified: 04 May 2023.
Why we need to know about logarithms
Logarithms were invented by John Napier, a Scottish mathematician, in the 17th century to simplify calculations.
Logarithms have many practical applications in various fields such as engineering, biology, chemistry, and economics. For example, they are used to measure the magnitude of earthquakes on the Richter scale, to calculate the pH of a solution in chemistry, and to model population growth in biology. In finance, logarithms are used to calculate compound interest and analyze stock market data. Understanding logarithms is essential for anyone studying advanced mathematics or science as they are powerful tools for solving complex problems.
The definition of logarithm
The logarithm of a positive number `y` is the power to which a base `a` must be raised to produce the number `y`. For example, if `2^4 = 16`, then `4 = log_2 16`.
If `y = a^x`, then `x = log_a y`,
where `a > 0`, `a!=1` and `y > 0`.
Logarithms and exponentials are inverse functions.
Properties of logarithms
Since logarithms and exponentials are inverse operations, the properties of logarithms are as follows:
(i) `log_a 1=0` (because `a^0 = 1`)
(ii) `log_a a^x = x` (because `a^x = a^x`)
(iii) `a^(log_a x) = x` (proof below)
Let `log_a x = y`, then `a^y = x`
Therefore `a^y = a^(log_a x) = x`
(iv) `log_a a = 1`
(from Property (iii), if we set `x = 1`).
Logarithm laws
We have the following logarithm laws:
(i) `log_a (xy) = log_a x + log_a y`
(ii) `log_a(x/y) = log_a x - log_a y`
(iii) `log_a x^n = n log_ax`
(iv) `log_a (1/x) = log_a x^(-1) = -log_a x`
Change of base
The change of base is useful across various fields: in electronics to measure signal strength with decibels; in computer programming for creating efficient algorithms; in compressing files, logarithms help measure how much info is squeezed, and changing bases helps compare different compression methods; in music, logarithms help find frequencies of notes, and the change of base formula lets us compare notes in different scales or tuning systems.
Most scientific calculators only directly calculate logarithms in base 10 and base e and have the following notations:
- the common logarithm (base 10): `log_10 x` or `log x`, and
- the natural logarithm (base e): `log_e x` or `ln x`.
Change of base formula
`log_a x = (log_b x)/(log_b a)`
can be used to find the logarithm of a number to any given base. For example,
`log_7 11 = (log 11)/(log 7)` or `(ln 11)/(ln 7)`
Here are some examples of how the logarithm laws and properties are used to solve various problems:
Example 1
Evaluate `log_2 16`.
We present two ways of going about evaluating this expression.
Solution (a)
Since `16=2^4`, then we can write
`log_2 16 = log_2 2^4`.
Now from Log Law (iii),
which says `log_a x^n = n log_ax`,
we have `log_2 2^4 = 4 log_2 2`.
Since Property (iv) says `log_a a = 1`,
we can write `4 log_2 2 = 4xx1 =4`.
Therefore,
`:. log_2 16 = 4`.
Solution (b)
Let's say that `x = log_2 16 `.
By definition, if `y = a^x`, then `x = log_a y`.
In the same way,
if `x = log_2 16`, then `16 = 2^x`.
If `16 = 2^x` and `16 = 2^4`,
then `2^x = 2^4`, so `x = 4`.
Now `x = 4` and `x = log_2 16 `,
Therefore,
`:. log_2 16 = 4`.
Example 2
Find the value of `x` that satisfies the equation
`3 log_10 x - 2 = 10`.
Solution
First, we move 2 to the right side:
`3 log_10 x = 10 + 2 = 12`
Next, by dividing both sides by 3,
we get `log_10 x = 12/3 = 4`
We use now the definition of logarithm:
if `log_10 x = 4`, then `x = 10^4 = 10000`
`:. x = 10^4 = 10000`
Example 3
Find `log_a 21` if `log_a 3 = p` and `log_a 7 = q`.
Solution
First, we write `21 = 3xx7`
Next, we use the Log Law (i),
`log_a (xy) = log_a x + log_a y`,
and write `log_a 21 = log_a (3xx7)`
`= log_a 3 + log_a 7`
But `log_a 3 = p` and `log_a 7 = q`,
therefore `log_a 21 = p + q`
`:.log_a 21 = p + q`
Example 4
Solve `log_2 12 = log_2 3 + log_2 x`.
Solution
We apply Log Law (i),
`log_a (xy) = log_a x + log_a y`,
to the right-hand side of the equation:
`log_2 3 + log_2 x = log_2 (3xxx)`
Now, our equation becomes
`log_2 12 = log_2 (3xxx)`,
therefore `12=3xxx`,
and by dividing both sides by 3,
we get `x = 4`.
`:. x = 4`
Example 5
Evaluate `log_6 4 - 2 log_6 12`.
Solution
First, we write
`2 log_6 12 = log_6 12^2`
`= log_6 144`
(we used Log Law (iii),
that says `log_a x^n = n log_ax`),
Now,
`log_6 4 - 2log_6 12`
`= log_6 4 - log_6 144`
`= log_6``4/144`
(we used Log Law (ii), which says
`log_a`` x/y = log_a x - log_a y`)
Now, we write
`log_6``4/144 = log_6``1/36`
`= log_6 ``1/6^2 = log_6 6^(-2)`
(we used the negative index law,
`a^(-m) = 1/a^m`)
To simplify `log_6 6^(-2)`, we use
Log Law (iii)) which is `log_a x^n = n log_ax`,
and Log Property (iv) which is `log_a a = 1`.
So,
`log_6 6^(-2) = -2 log_6 6 = -2xx1 = -2`
Therefore,
`:. log_6 4 - 2 log_6 12 = -2`
Example 6
`A = 100 - 50 log_10 (t+1)` is the amount of a substance at time t. We are given various amounts and we need to find when those amounts occur. Change the subject to t.
Solution
`A = 100 - 50 log_10 (t+1)`
(subtract `100` on both sides)
`A - 100 = -50 log _10 (t+1)`
(divide by `-50` on both sides)
`(A-100)/-50 = log_10 (t+1)`
Now, if `(A-100)/-50 = log_10 (t+1)`,
then `t+1 = 10^ ((A-100)/-50)`
(we used the definition of log which says,
if `y = a^x`, then `x = log_a y`)
Now, in `t+1 = 10^ ((A-100)/-50)`, we subtract 1 on both sides:
`t = 10^ ((A-100)/-50)-1`
Rewrite `(A-100)/-50` by multiplying both, the numerator and denominator, by `-1`:
`(A-100)/-50 = (-1xx(A-100))/(-1xx(-50))`
`= (-A+100)/50 = -A/50 + 100/50 `
`=-A/50+2`
Therefore,
`:. t = 10^ (-A/50+2) - 1`