Laws of Logarithms
Every question with a fully worked solution.
- Laws of Logarithms - Vidoe - Logarithms : Simplifying and Expanding Watch
Theory
The log laws turn multiplication into addition: \(\log_a(xy) = \log_a x + \log_a y\), \(\log_a \dfrac{x}{y} = \log_a x - \log_a y\) and \(\log_a x^n = n\log_a x\). This Year 10A maths page uses them to evaluate and simplify, write a single log, work with given values, solve log equations and change the base.
A logarithm is an index, so the index laws become log laws. Multiplying numbers adds their logs, dividing subtracts them, and a power becomes a multiplier in front.
The laws only work when every log has the same base and every number inside a log is positive.
Expanding splits one log into several; condensing combines several into a single log. To combine a plain number, first write it as a log: \(1 = \log_a a\), \(2 = \log_a a^2\).
The change of base rule lets a calculator, which works in base \(10\), evaluate a log to any base.
This prepares you for Mathematical Methods and Specialist Mathematics in Years 11 and 12.
For \(a \gt 0\), \(a \ne 1\) and \(x, y \gt 0\):
Special results:
Change of base:
How to simplify or evaluate
- Powers first: move any multiplier in front of a log into the index: \(3\log_2 5 = \log_2 125\). Write plain numbers as logs of the base.
- Combine sums into products and differences into quotients, giving a single log.
- Evaluate by writing the number inside as a power of the base. Decimals: write them as fractions first, \(0.04 = \dfrac{1}{25}\).
- A fraction of two logs with the same base: write both numbers as powers of the same prime, then cancel the common log.
Other question types
- Given values: factorise the number into the given primes (\(54 = 2 \times 3^3\), \(0.16 = \dfrac{2^2}{5^2}\)), expand, then substitute.
- Only \(\log_{10} 2\) given: build the number from powers of \(10\) and powers of \(2\), for example \(25 = \dfrac{100}{2^2}\) or \(0.4 = \dfrac{2^2}{10}\).
- Log equations: condense each side into one log, change to index form, solve, then reject any answer that makes a number inside a log zero or negative.
- Change of base: divide base-\(10\) logs. In a product of logs with linked bases, change all of them and cancel the matching terms.
(i) Evaluate \(\log_6 9 + \log_6 4\).
(ii) Evaluate \(\log_2 40 - \log_2 5\).
(iii) Simplify \(\dfrac{\log_5 27}{\log_5 9}\).
(iv) Write \(2\log_5 3 + 1\) as a single logarithm.
(i) Product law, then recognise a power of \(6\).
| \(\log_6 9 + \log_6 4\) | \(=\) | \(\log_6 36\) |
| \(\) | \(=\) | \(\mathbf{2}\) |
(ii) Quotient law.
| \(\log_2 40 - \log_2 5\) | \(=\) | \(\log_2 \dfrac{40}{5} = \log_2 8\) |
| \(\) | \(=\) | \(\mathbf{3}\) |
(iii) There is no law for dividing two logs. Write both numbers as powers of \(3\) and use the power law.
| \(\dfrac{\log_5 27}{\log_5 9}\) | \(=\) | \(\dfrac{3\log_5 3}{2\log_5 3}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{3}{2}}\) |
(iv) Write \(1\) as \(\log_5 5\), move the \(2\) into the index, then combine.
| \(2\log_5 3 + 1\) | \(=\) | \(\log_5 3^2 + \log_5 5\) |
| \(\) | \(=\) | \(\log_5(9 \times 5)\) |
| \(\) | \(=\) | \(\mathbf{\log_5 45}\) |
(i) If \(\log_a 3 = m\) and \(\log_a 5 = n\), express \(\log_a 75\) and \(\log_a 0.6\) in terms of \(m\) and \(n\).
(ii) Given \(\log_{10} 2 \approx 0.301\) and \(\log_{10} 3 \approx 0.477\), find \(\log_{10} 4.5\) without a calculator.
(iii) If \(\log_a x = 2\) and \(\log_a y = 5\), find \(\log_a\left(\dfrac{x^3}{\sqrt y}\right)\).
(i) Write each number using only \(3\)s and \(5\)s: \(75 = 3 \times 5^2\) and \(0.6 = \dfrac{3}{5}\).
| \(\log_a 75\) | \(=\) | \(\log_a 3 + 2\log_a 5 = \boldsymbol{m + 2n}\) |
| \(\log_a 0.6\) | \(=\) | \(\log_a 3 - \log_a 5 = \boldsymbol{m - n}\) |
(ii) \(4.5 = \dfrac{9}{2} = \dfrac{3^2}{2}\).
| \(\log_{10} 4.5\) | \(=\) | \(2\log_{10} 3 - \log_{10} 2\) |
| \(\) | \(\approx\) | \(2(0.477) - 0.301\) |
| \(\) | \(=\) | \(\mathbf{0.653}\) |
(iii) Expand, writing \(\sqrt y = y^{\genfrac{}{}{}{1}{1}{2}}\), then substitute.
| \(\log_a\left(\dfrac{x^3}{\sqrt y}\right)\) | \(=\) | \(3\log_a x - \dfrac{1}{2}\log_a y\) |
| \(\) | \(=\) | \(3(2) - \dfrac{1}{2}(5)\) |
| \(\) | \(=\) | \(\mathbf{3.5}\) |
(i) \(\log_3 x + \log_3(x + 8) = 2\)
(ii) \(\log_2(x + 6) - \log_2(x - 1) = 3\)
(i) Condense to one log, change to index form, then solve.
| \(\log_3\big(x(x + 8)\big)\) | \(=\) | \(2\) |
| \(x^2 + 8x\) | \(=\) | \(3^2\) |
| \(x^2 + 8x - 9\) | \(=\) | \(0\) |
| \((x + 9)(x - 1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-9 \text{ or } 1\) |
\(x = -9\) would need \(\log_3(-9)\), which does not exist, so reject it. \(x = \mathbf{1}\). Check: \(\log_3 1 + \log_3 9 = 0 + 2 = 2\).
(ii) Quotient law.
| \(\log_2 \dfrac{x + 6}{x - 1}\) | \(=\) | \(3\) |
| \(\dfrac{x + 6}{x - 1}\) | \(=\) | \(2^3 = 8\) |
| \(x + 6\) | \(=\) | \(8x - 8\) |
| \(x\) | \(=\) | \(\mathbf{2}\) |
Check: \(\log_2 8 - \log_2 1 = 3 - 0 = 3\), and both numbers inside the logs are positive.
(i) Evaluate \(\log_3 7\), correct to three decimal places.
(ii) Show that \(\log_2 5 \times \log_5 32 = 5\).
(i) A calculator works in base \(10\), so change the base.
| \(\log_3 7\) | \(=\) | \(\dfrac{\log_{10} 7}{\log_{10} 3}\) |
| \(\) | \(=\) | \(1.7712\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.771}\) |
Check: \(3^{1.771} \approx 7\). It is between \(1\) and \(2\), as \(3 \lt 7 \lt 9\) says it must be.
(ii) Change both logs to base \(10\); the \(\log_{10} 5\) cancels.
| \(\log_2 5 \times \log_5 32\) | \(=\) | \(\dfrac{\log_{10} 5}{\log_{10} 2} \times \dfrac{\log_{10} 32}{\log_{10} 5}\) |
| \(\) | \(=\) | \(\dfrac{\log_{10} 32}{\log_{10} 2} = \log_2 32\) |
| \(\) | \(=\) | \(\mathbf{5}\) |
Common pitfalls
Frequently asked questions
What are the three laws of logarithms?
The product law \(\log_a(xy) = \log_a x + \log_a y\), the quotient law \(\log_a \dfrac{x}{y} = \log_a x - \log_a y\) and the power law \(\log_a x^n = n\log_a x\). All logs must have the same base.
How do you write an expression as a single logarithm?
Move any number in front of a log into the index, write any plain number as a log of the base, then combine sums as products and differences as quotients.
Can you add logs with different bases?
Not with the log laws. Either evaluate each log separately or use the change of base rule to write them all in the same base first.
How do you calculate a log with a different base on a calculator?
Use the change of base rule: \(\log_a b = \dfrac{\log_{10} b}{\log_{10} a}\). For example \(\log_3 7 \approx 1.771\).
Why do I have to check answers to log equations?
Combining logs can create a solution that makes a number inside one of the original logs zero or negative. Those logs do not exist, so that solution must be rejected.