Exponents (powers) are a shorthand for repeated multiplication, and logarithms are the way to undo them. This note covers the laws of exponents for integer powers and the first steps with base 10 and natural logarithms, evaluated with technology. You need these ideas for growth and decay, pH, decibels, and later work on functions and calculus.
Key ideas
In \(a^{n}\), the number \(a\) is the base and \(n\) is the exponent. For a positive whole number \(n\), \(a^n\) means \(a\) multiplied by itself \(n\) times. We extend this to zero and negative exponents so that the same laws keep working.
The laws of exponents for integer exponents (with \(a \ne 0\) and \(b \ne 0\)):
- Multiplying powers of the same base: add exponents, \(a^m \times a^n = a^{m+n}\).
- Dividing powers of the same base: subtract exponents, \(\dfrac{a^m}{a^n} = a^{m-n}\).
- Power of a power: multiply exponents, \((a^m)^n = a^{mn}\).
- Power of a product: \((ab)^n = a^n b^n\), and of a quotient \(\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}\).
- Zero exponent: \(a^0 = 1\).
- Negative exponent: \(a^{-n} = \dfrac{1}{a^n}\). A negative exponent means a reciprocal, not a negative number.
When you simplify an expression with several letters, deal with the numbers first, then each letter in turn. Brackets with a power outside apply the power to every factor inside, including the number at the front.
Logarithms. A logarithm answers the question "what power do I need?". The statement
\[a^{x} = b \iff x = \log_a b \quad (a \gt 0,\ a \ne 1,\ b \gt 0)\]
says that \(\log_a b\) is the power to which \(a\) must be raised to give \(b\). For example, \(\log_{10} 1000 = 3\) because \(10^3 = 1000\), and \(\log_{10} 0.01 = -2\) because \(10^{-2} = 0.01\).
At this stage you work with two bases. The common logarithm, base 10, is written \(\log x\) (or \(\log_{10} x\)). The natural logarithm has base \(e \approx 2.718\) and is written \(\ln x\). Logarithms and exponentials with the same base are inverses, so \(10^{\log x} = x\), \(\log 10^{x} = x\), \(e^{\ln x} = x\) and \(\ln e^{x} = x\). Also \(\log 1 = 0\) and \(\ln e = 1\). You cannot take the logarithm of zero or of a negative number, because no power of a positive base gives such a result.
Use your GDC for numerical values, with the "log" key for base 10 and the "ln" key for base \(e\). To solve a simple equation of the form \(10^{x} = b\) or \(e^{x} = b\), take the logarithm of both sides in the matching base, which gives \(x = \log b\) or \(x = \ln b\). If the exponent has a coefficient or other terms, first isolate the power, then take the logarithm.
Why logarithms matter. Many real quantities grow or shrink by a fixed factor in equal steps: a bacteria colony, a cooling drink, a loudness scale. Their formulae contain the unknown in the exponent, such as \(500 \times 1.3^{t} = 2000\). Ordinary algebra cannot bring the exponent down, but a logarithm can, because it turns "which power?" into a direct question. Logarithmic scales also compress huge ranges into small numbers. On the pH scale, each step of 1 means the concentration changes by a factor of 10.
Checking by estimation. Before you use a calculator, estimate. Since \(10^2 = 100\) and \(10^3 = 1000\), the value of \(\log 250\) must lie between 2 and 3. Because \(e^3 \approx 20\) and \(e^4 \approx 55\), the value of \(\ln 40\) must lie between 3 and 4. A quick estimate like this catches a mistyped key or a wrong base.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(a^m \times a^n = a^{m+n}\) | Same base, multiplying. |
| \(\dfrac{a^m}{a^n} = a^{m-n}\) | Same base, dividing. |
| \((a^m)^n = a^{mn}\) | Power of a power. |
| \(a^0 = 1\), \(a^{-n} = \dfrac{1}{a^n}\) | Zero and negative exponents. |
| \(a^x = b \iff x = \log_a b\) | Definition of a logarithm. |
| \(\log x\), \(\ln x\) | Logarithms to base 10 and base \(e\), evaluated with technology. |
| \(e^{\ln x} = x\), \(\ln e^x = x\) | Natural logarithm and exponential undo each other. |
Worked examples
Example 1 Simplify \(\dfrac{(2x^3y^{-2})^3}{4x^2y^{-1}}\), giving your answer with positive exponents.
Numerator: \((2x^3y^{-2})^3 = 2^3 \times x^{9} \times y^{-6} = 8x^9y^{-6}\).
Divide by the denominator: \(\dfrac{8}{4} = 2\), \(x^{9-2} = x^7\), \(y^{-6-(-1)} = y^{-5}\).
So the expression is \(2x^7y^{-5}\). Write \(y^{-5}\) as \(\dfrac{1}{y^5}\).
Answer: \(\dfrac{2x^7}{y^5}\).
Example 2 (a) Write down the exact values of \(\log 1000\) and \(\ln e^{5}\). (b) The pH of a solution is \(-\log[\mathrm{H}^{+}]\). Find the pH when \([\mathrm{H}^{+}] = 3.2 \times 10^{-5}\) mol per litre, to 3 significant figures.
(a) \(10^3 = 1000\), so \(\log 1000 = 3\). Since \(\ln\) and \(e^x\) undo each other, \(\ln e^5 = 5\).
(b) Use the "log" key on your GDC: \(\log(3.2 \times 10^{-5}) = -4.49485\ldots\)
pH \(= -(-4.49485\ldots) = 4.49485\ldots\)
Answer: (a) 3 and 5. (b) pH \(\approx 4.49\).
Example 3 Solve (a) \(10^{x} = 250\), (b) \(e^{2x} = 40\), (c) \(2 \times 10^{x} = 7\). Give each answer to 3 significant figures.
(a) \(x = \log 250 = 2.39794\ldots\), so \(x \approx 2.40\).
(b) The exponent is \(2x\), so \(2x = \ln 40 = 3.68888\ldots\), giving \(x = 1.84444\ldots \approx 1.84\).
(c) Isolate the power first: \(10^{x} = 3.5\). Then \(x = \log 3.5 = 0.54407\ldots \approx 0.544\).
Answer: (a) 2.40. (b) 1.84. (c) 0.544.
Common errors
- Common error: multiplying the bases when multiplying powers. \(2^3 \times 2^4 = 2^7\), not \(4^7\).
- Common error: thinking \(a^{-n}\) is negative. \(2^{-3} = \tfrac{1}{8}\), which is positive.
- Common error: forgetting to raise the numerical coefficient to the power. \((2x^3)^3 = 8x^9\), not \(2x^9\).
- Common error: taking the logarithm before isolating the power. In \(2 \times 10^{x} = 7\), divide by 2 first.
- Common error: mixing up \(\log\) (base 10) and \(\ln\) (base \(e\)) when the equation uses the other base.
Exam tip: When you solve an exponential equation, write down the logarithm statement exactly (for example \(x = \log 250\)) before you calculate. Keep full calculator values and round only at the end.
Check yourself
- Simplify \((a^3b^{-1})^{-2} \times a^5b\), giving your answer with positive exponents.
Answer
\((a^3b^{-1})^{-2} = a^{-6}b^{2}\). Multiply by \(a^5b\): \(a^{-6+5}b^{2+1} = a^{-1}b^{3} = \dfrac{b^3}{a}\). - Write down the value of \(\ln\left(e^{-3}\right)\) and of \(\log 0.001\).
Answer
\(\ln e^{-3} = -3\). \(0.001 = 10^{-3}\), so \(\log 0.001 = -3\). - Solve \(e^{x} = 7.5\) to 3 significant figures.
Answer
\(x = \ln 7.5 = 2.0149\ldots \approx 2.01\). - Solve \(10^{x} = 0.04\) to 3 significant figures.
Answer
\(x = \log 0.04 = -1.39794\ldots \approx -1.40\).