This note extends exponents to fractional powers and develops the laws of logarithms, including change of base. These tools let you simplify expressions with roots and solve equations where the unknown is in an exponent or inside a logarithm. They underpin growth and decay models, and calculus with exponential and logarithmic functions.
Key ideas
Rational exponents. A fractional power means a root combined with a power:
\[a^{\frac{1}{n}} = \sqrt[n]{a}, \qquad a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^{m}}\]
The denominator is the root and the numerator is the power. It is usually easiest to take the root first, because that keeps the numbers small. For example, \(27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9\). A negative rational exponent still means a reciprocal, so \(16^{-3/4} = \dfrac{1}{16^{3/4}} = \dfrac{1}{8}\). All the laws of exponents from integer powers still apply.
Laws of logarithms. For \(a \gt 0\), \(a \ne 1\), and positive \(x\) and \(y\):
- \(\log_a(xy) = \log_a x + \log_a y\) (a product becomes a sum).
- \(\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y\) (a quotient becomes a difference).
- \(\log_a(x^{m}) = m\log_a x\) (a power comes down as a multiplier). This works for any real \(m\), including fractions.
- \(\log_a a = 1\), \(\log_a 1 = 0\), and \(a^{\log_a x} = x\).
These laws come directly from the laws of exponents. The sum law, for example, mirrors \(a^m \times a^n = a^{m+n}\). Notice that \(\log_a(x + y)\) has no simple expansion. You cannot split a logarithm of a sum.
Change of base. To evaluate a logarithm in a base your calculator does not have, use
\[\log_a x = \frac{\log_b x}{\log_b a}\]
where \(b\) is any convenient base, usually 10 or \(e\). For example, \(\log_5 40 = \dfrac{\ln 40}{\ln 5} = 2.29\) to 3 s.f.
Solving exponential equations. When the unknown is in the exponent and the two sides cannot be written with the same base, take logarithms of both sides (or convert directly using the definition \(a^x = b \iff x = \log_a b\)), then use the power law to bring the exponent down. Isolate the exponential term first.
Solving equations with logarithms. Combine logarithms into a single logarithm, rewrite the equation in exponent form, and solve. Then always check the answers in the original equation. A logarithm can only accept positive inputs, so any solution that makes an input zero or negative must be rejected.
Using logarithms in context. Suppose an investment of 1000 grows by 6% each year and you want to know when it reaches 2500. The equation is \(1000 \times 1.06^{t} = 2500\), so \(1.06^{t} = 2.5\). Taking logarithms gives \(t \ln 1.06 = \ln 2.5\), so \(t = \dfrac{\ln 2.5}{\ln 1.06} = 15.7\) years. The power law is what lets you move \(t\) out of the exponent. It does not matter whether you use \(\log\) or \(\ln\), as long as you use the same one on both sides.
Working with roots. To handle expressions with roots, rewrite them as rational exponents first: \(\sqrt{x} = x^{1/2}\) and \(\dfrac{1}{\sqrt[3]{x}} = x^{-1/3}\). Then use the laws of exponents to combine terms. This is especially useful when you simplify products and quotients of powers of a letter, and it prepares you for differentiating such expressions later in the course.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(a^{m/n} = \left(\sqrt[n]{a}\right)^m\) | Rational exponent: root first, then power. |
| \(\log_a(xy) = \log_a x + \log_a y\) | Product law. |
| \(\log_a\!\left(\dfrac{x}{y}\right) = \log_a x - \log_a y\) | Quotient law. |
| \(\log_a(x^m) = m\log_a x\) | Power law. Used to bring an exponent down. |
| \(\log_a x = \dfrac{\log_b x}{\log_b a}\) | Change of base, for evaluating with a calculator. |
| \(a^x = b \iff x = \log_a b\) | Link between exponential and logarithmic form. |
Worked examples
Example 1 (a) Evaluate \(125^{-2/3}\). (b) Simplify \((8x^6)^{-2/3}\), giving your answer with a positive exponent. (c) Write \(2\log 3 + \log 20 - \log 18\) as a single logarithm and hence evaluate it.
(a) \(125^{-2/3} = \dfrac{1}{125^{2/3}} = \dfrac{1}{(\sqrt[3]{125})^2} = \dfrac{1}{5^2} = \dfrac{1}{25}\).
(b) Apply the power to each factor: \(8^{-2/3} \times (x^6)^{-2/3} = \dfrac{1}{(\sqrt[3]{8})^2} \times x^{-4} = \dfrac{1}{4} \times x^{-4} = \dfrac{1}{4x^4}\).
(c) \(2\log 3 = \log 3^2 = \log 9\). Then \(\log 9 + \log 20 - \log 18 = \log\dfrac{9 \times 20}{18} = \log 10 = 1\).
Answer: (a) \(\tfrac{1}{25}\). (b) \(\dfrac{1}{4x^4}\). (c) 1.
Example 2 Solve \(3^{2x-1} = 50\), giving your answer to 3 significant figures. State also the value of \(\log_3 50\) to 3 significant figures.
By the definition of a logarithm, \(2x - 1 = \log_3 50\).
Use change of base: \(\log_3 50 = \dfrac{\ln 50}{\ln 3} = \dfrac{3.91202\ldots}{1.09861\ldots} = 3.56088\ldots\)
So \(2x = 4.56088\ldots\) and \(x = 2.28044\ldots\)
Answer: \(x \approx 2.28\) and \(\log_3 50 \approx 3.56\).
Example 3 Solve \(\log_2(x+3) + \log_2(x-3) = 4\).
Combine using the product law: \(\log_2\big((x+3)(x-3)\big) = 4\), so \(\log_2(x^2 - 9) = 4\).
Convert to exponent form: \(x^2 - 9 = 2^4 = 16\), so \(x^2 = 25\) and \(x = 5\) or \(x = -5\).
Check the inputs. For \(x = -5\), \(x - 3 = -8\), which is negative, so the logarithm is not defined. Reject \(x = -5\).
For \(x = 5\): \(\log_2 8 + \log_2 2 = 3 + 1 = 4\). Correct.
Answer: \(x = 5\).
Common errors
- Common error: writing \(\log(x+y) = \log x + \log y\). The product law applies to \(\log(xy)\), not to a sum.
- Common error: writing \(\dfrac{\log x}{\log y} = \log\left(\dfrac{x}{y}\right)\). The quotient law is for \(\log x - \log y\). A quotient of two logarithms is a change of base instead.
- Common error: evaluating \(a^{m/n}\) as \(a^{m}\) divided by \(n\). The denominator is a root.
- Common error: forgetting to reject a solution that makes a logarithm's input negative or zero.
- Common error: bringing down only part of an exponent, for example \(\log 3^{2x-1} = 2x - 1\log 3\). The whole exponent multiplies: \((2x-1)\log 3\).
Exam tip: After solving a logarithmic equation, substitute each answer back to confirm every logarithm has a positive input, and write a line saying which solution you reject and why.
Check yourself
- Evaluate \(32^{3/5}\).
Answer
\(\sqrt[5]{32} = 2\), so \(32^{3/5} = 2^3 = 8\). - Write \(\log 5 + \log 8 - \log 4\) as a single number.
Answer
\(\log\dfrac{5 \times 8}{4} = \log 10 = 1\). - Solve \(7^{x} = 20\) to 3 significant figures.
Answer
\(x = \log_7 20 = \dfrac{\ln 20}{\ln 7} = \dfrac{2.99573\ldots}{1.94591\ldots} = 1.5395\ldots \approx 1.54\). - Solve \(\ln x + \ln(x-2) = \ln 15\).
Answer
\(\ln\big(x(x-2)\big) = \ln 15\), so \(x^2 - 2x = 15\), giving \(x^2 - 2x - 15 = 0\), \((x-5)(x+3) = 0\). So \(x = 5\) or \(x = -3\). For \(x = -3\) the input of \(\ln x\) is negative, so reject it. Answer: \(x = 5\).