What a logarithm is
A logarithm answers the question "what power?". If you know that \(2^3 = 8\), then the logarithm turns that around and asks: to what power must I raise \(2\) to get \(8\)? The answer is \(3\), written \(\log_2 8 = 3\).
The formal link between powers and logs is the single most important line in this note:
Read it both ways. The left side is the "power" form; the right side is the "log" form. They say exactly the same thing. Here \(b\) is the base (a positive number, not \(1\)), \(x\) is the number you are taking the log of (which must be positive), and \(y\) is the power, the answer the logarithm gives you.
- Logarithm
- \(\log_b x\) is the power to which the base \(b\) must be raised to produce \(x\). For example \(\log_{10} 1000 = 3\) because \(10^3 = 1000\).
Where this is assessed
Converting between power and log form, and applying the laws, is Criterion A and Criterion C. Solving a real exponential model, such as growth or decay, and interpreting the result is Criterion D.
The laws of logs
Because logs are really indices in disguise, they inherit the index laws. Three are essential. In each, the base \(b\) is the same throughout.
| Law | Rule | In words |
|---|---|---|
| Product | \(\log_b(mn) = \log_b m + \log_b n\) | A log of a product is the sum of the logs. |
| Quotient | \(\log_b\!\left(\dfrac{m}{n}\right) = \log_b m - \log_b n\) | A log of a quotient is the difference of the logs. |
| Power | \(\log_b(m^k) = k\log_b m\) | A power inside a log comes out as a multiplier. |
Two special values are worth memorising: \(\log_b 1 = 0\) (because \(b^0 = 1\)) and \(\log_b b = 1\) (because \(b^1 = b\)).
There is no law for adding inside
The product law is about \(\log(mn)\), not \(\log(m + n)\). There is no rule that simplifies \(\log_b(m + n)\), so never rewrite it as \(\log_b m + \log_b n\). That single false move is a very common trap.
Change of base
Most calculators only offer \(\log\) (base \(10\)) and \(\ln\) (base \(e\)). To evaluate a log in any other base, use the change of base rule:
You may pick any base \(c\) for the right side, so choose one your calculator has, usually \(10\). For example, \(\log_2 20 = \dfrac{\log_{10} 20}{\log_{10} 2}\), which a calculator turns into a decimal.
Solving log and exponential equations
The definition and the laws let you unlock equations where the unknown is trapped inside a log or up in a power. The trick is to move between the two forms.
Solve \(\log_2 x + \log_2 3 = \log_2 15\).
When the unknown is in the power, take a log of both sides and let the power law bring it down. On a calculator you then apply change of base.
Solve \(2^x = 20\), giving your answer to 2 decimal places.
Sense-check the size
Before trusting a decimal answer, bracket it. Since \(2^4 = 16\) and \(2^5 = 32\), and \(20\) sits between them, the answer must be between \(4\) and \(5\). \(4.32\) fits, so it is believable.
Check yourself
Work through each, then reveal the full answer.
1. Evaluate \(\log_5 125\). +
Ask "5 to what power gives 125?" Since \(5^3 = 125\), the answer is \(\log_5 125 = 3\).
2. Write \(\log a + 2\log b\) as a single logarithm. +
First use the power law on the second term: \(2\log b = \log b^2\). Then use the product law on the sum: \(\log a + \log b^2 = \log(ab^2)\). So the answer is \(\log(ab^2)\).
3. Solve \(3^x = 30\), to 2 decimal places. +
Rewrite in log form: \(x = \log_3 30\). Change base: \(x = \dfrac{\log_{10} 30}{\log_{10} 3} = \dfrac{1.47712}{0.47712} = 3.0959\ldots\) Since \(3^3 = 27\) and \(3^4 = 81\), a value near \(3.1\) is sensible. So \(x \approx 3.10\).
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