Introducing Logarithms
Theory
A logarithm is an index: \(\log_a y\) is the power you raise \(a\) to in order to get \(y\). This Year 10A maths page shows how to switch between index form and logarithm form, evaluate logs of whole numbers, fractions and surds, estimate a log, solve simple log equations and use logarithmic scales.
The logarithm of a number \(y\) to the base \(a\) is the index that \(a\) must be raised to in order to give \(y\). So \(a^x = y\) and \(\log_a y = x\) say exactly the same thing. For example \(10^3 = 1000\), so \(\log_{10} 1000 = 3\).
The base must be positive and not equal to 1. Because \(a^x\) is always positive, you can only take the log of a positive number: \(\log_2 0\) and \(\log_2(-4)\) do not exist.
To evaluate a log without a calculator, write the base and the number as powers of the same prime and compare indices. Fractions give negative indices and roots give fractional indices.
A logarithmic scale (pH, decibels, earthquake magnitude) records the log of a quantity. Each step of \(1\) on a base-\(10\) scale means the quantity is multiplied by \(10\).
This prepares you for Mathematical Methods and Specialist Mathematics in Years 11 and 12.
Results that follow straight from the definition:
| Number | As a power of 2 | \(\log_2\) of it |
|---|---|---|
| \(16\) | \(2^4\) | \(4\) |
| \(\dfrac{1}{8}\) | \(2^{-3}\) | \(-3\) |
| \(\sqrt2\) | \(2^{\genfrac{}{}{}{1}{1}{2}}\) | \(\dfrac{1}{2}\) |
| \(\sqrt[5]{2}\) | \(2^{\genfrac{}{}{}{1}{1}{5}}\) | \(\dfrac{1}{5}\) |
| \(2\sqrt2\) | \(2^{\genfrac{}{}{}{1}{3}{2}}\) | \(\dfrac{3}{2}\) |
How to evaluate a logarithm
- Let the log equal \(x\): \(\log_a y = x\).
- Rewrite in index form: \(a^x = y\).
- Express both sides as powers of the same prime, using negative indices for fractions and fractional indices for roots.
- Equate the indices and solve for \(x\).
Solving log equations and other problems
- Unknown inside the log: change to index form, solve, then check that the number inside the log is positive.
- Unknown base: change to index form and take the root. Reject any base that is negative, zero or \(1\).
- A log squared: let \(m = \log_a x\), solve the quadratic in \(m\), then change each answer back.
- A log of a log: undo the outer log first, then the inner one.
- Two equations: change each to index form and substitute, for example \(x = 2^y\) into the second equation. Combine powers with the index laws and equate indices.
- Given values: if \(\log_3 p = 2\) then \(p = 9\). Find each pronumeral first, then evaluate the new log.
- Estimating: find consecutive powers of the base either side of the number (Example 2).
(i) \(\log_{16} 8\)
(ii) \(\log_{0.2} 125\)
(iii) \(\log_{\sqrt5} 25\)
(iv) \(\log_8 \dfrac{1}{4\sqrt2}\)
Write the base and the number as powers of the same prime, then compare indices.
(i) \(16 = 2^4\) and \(8 = 2^3\).
| \(\text{Let } \log_{16} 8\) | \(=\) | \(x\) |
| \(16^x\) | \(=\) | \(8\) |
| \(2^{4x}\) | \(=\) | \(2^3\) |
| \(x\) | \(=\) | \(\mathbf{\dfrac{3}{4}}\) |
(ii) \(0.2 = \dfrac{1}{5} = 5^{-1}\) and \(125 = 5^3\).
| \((5^{-1})^x\) | \(=\) | \(5^3\) |
| \(-x\) | \(=\) | \(3\) |
| \(x\) | \(=\) | \(\mathbf{-3}\) |
(iii) \(\sqrt5 = 5^{\genfrac{}{}{}{1}{1}{2}}\) and \(25 = 5^2\).
| \(5^{\genfrac{}{}{}{1}{x}{2}}\) | \(=\) | \(5^2\) |
| \(\dfrac{x}{2}\) | \(=\) | \(2\) |
| \(x\) | \(=\) | \(\mathbf{4}\) |
(iv) \(8 = 2^3\) and \(4\sqrt2 = 2^2\times2^{\genfrac{}{}{}{1}{1}{2}} = 2^{\genfrac{}{}{}{1}{5}{2}}\), so \(\dfrac{1}{4\sqrt2} = 2^{-\genfrac{}{}{}{1}{5}{2}}\).
| \(2^{3x}\) | \(=\) | \(2^{-\genfrac{}{}{}{1}{5}{2}}\) |
| \(3x\) | \(=\) | \(-\dfrac{5}{2}\) |
| \(x\) | \(=\) | \(\mathbf{-\dfrac{5}{6}}\) |
A number less than \(1\) with a base greater than \(1\) always gives a negative log.
(i) Write \(5^{-2} = \dfrac{1}{25}\) in logarithm form, and \(\log_{16} 2 = \dfrac{1}{4}\) in index form.
(ii) Between which two consecutive whole numbers does \(\log_3 100\) lie?
(iii) Solve \(\log_x 81 = 4\).
(i) The base stays the base and the index becomes the value of the log.
| \(5^{-2} = \dfrac{1}{25}\) | \(\Leftrightarrow\) | \(\log_5 \dfrac{1}{25} = -2\) |
| \(\log_{16} 2 = \dfrac{1}{4}\) | \(\Leftrightarrow\) | \(16^{\genfrac{}{}{}{1}{1}{4}} = 2\) |
(ii) Find the powers of \(3\) either side of \(100\): \(3^4 = 81\) and \(3^5 = 243\).
| \(81 \lt 100\) | \(\lt\) | \(243\) |
| \(\log_3 81 \lt \log_3 100\) | \(\lt\) | \(\log_3 243\) |
| \(4 \lt \log_3 100\) | \(\lt\) | \(5\) |
So \(\log_3 100\) lies between \(\mathbf{4}\) and \(\mathbf{5}\).
(iii) Write in index form and take the fourth root.
| \(x^4\) | \(=\) | \(81\) |
| \(x\) | \(=\) | \(3 \text{ or } -3\) |
A base must be positive and not \(1\), so reject \(-3\): \(x = \mathbf{3}\).
(i) \(\log_3(x^2 - 2x) = 1\)
(ii) \((\log_2 x)^2 - 5\log_2 x + 6 = 0\)
(iii) \(\log_3(\log_2 x) = 1\)
(i) Change to index form, then solve the quadratic.
| \(x^2 - 2x\) | \(=\) | \(3^1\) |
| \(x^2 - 2x - 3\) | \(=\) | \(0\) |
| \((x - 3)(x + 1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\mathbf{3} \text{ or } \mathbf{-1}\) |
Check the number inside the log: \(3^2 - 2(3) = 3\) and \((-1)^2 - 2(-1) = 3\). Both are positive, so both solutions stand.
(ii) The equation is a quadratic in \(\log_2 x\). Let \(m = \log_2 x\).
| \(m^2 - 5m + 6\) | \(=\) | \(0\) |
| \((m - 2)(m - 3)\) | \(=\) | \(0\) |
| \(\log_2 x = 2 \text{ or } \log_2 x\) | \(=\) | \(3\) |
| \(x\) | \(=\) | \(\mathbf{4} \text{ or } \mathbf{8}\) |
(iii) Undo the outside log first, then the inside one.
| \(\log_2 x\) | \(=\) | \(3^1 = 3\) |
| \(x\) | \(=\) | \(2^3 = \mathbf{8}\) |
Check: \(\log_2 8 = 3\) and \(\log_3 3 = 1\).
(i) A tremor has an amplitude \(5000\) times \(A_0\). Find its magnitude, correct to one decimal place.
(ii) How many times as large is the amplitude of a magnitude \(6.2\) quake as that of a magnitude \(4.7\) quake? Answer to the nearest whole number.
(i) Substitute \(\dfrac{A}{A_0} = 5000\).
| \(M\) | \(=\) | \(\log_{10} 5000\) |
| \(\) | \(=\) | \(3.6989\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{3.7}\) |
(ii) In index form \(\dfrac{A}{A_0} = 10^M\). Divide the two amplitudes and use the index laws.
| \(\dfrac{A_1}{A_2}\) | \(=\) | \(\dfrac{10^{6.2}A_0}{10^{4.7}A_0}\) |
| \(\) | \(=\) | \(10^{1.5}\) |
| \(\) | \(=\) | \(31.6227\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{32} \text{ times}\) |
On a log scale, a difference of \(1\) means \(10\) times as large. A difference of \(1.5\) means \(10^{1.5}\) times, not \(15\) times.
Common pitfalls
Frequently asked questions
What is a logarithm in simple terms?
A logarithm is an index. \(\log_a y\) is the power you raise \(a\) to in order to get \(y\). For example \(\log_2 8 = 3\), because \(2^3 = 8\).
How do you convert between index form and log form?
\(a^x = y\) is the same as \(\log_a y = x\). The base stays the base and the index becomes the value of the logarithm.
Can a logarithm be negative?
Yes. The value of a log can be negative, for example \(\log_{10} 0.01 = -2\). But the number you take the log of must be positive.
Why can't you take the log of a negative number?
A positive base raised to any power is always positive, so no power can give zero or a negative number. That means the log of zero or a negative number does not exist.
How do you find a log without a calculator?
Write the base and the number as powers of the same prime, using negative indices for fractions and fractional indices for roots, then equate the indices.
What does a logarithmic scale mean?
It records the log of a quantity instead of the quantity. On a base 10 scale, each increase of 1 means the quantity is 10 times as large, so a difference of 3 means 1000 times.