Learning objectives
Write and calculate with numbers in the form a × 10ᵏ, where a is at least 1 and less than 10 and k is an integer
Apply the laws of exponents with integer exponents and evaluate base 10 and base e logarithms using the equivalence a = bˣ ⟺ x = log_b a
Round to a given number of decimal places or significant figures and state the upper and lower bounds of a rounded quantity
Calculate and interpret percentage error, and use one-significant-figure estimates to check that a calculator answer is sensible
Key definitions
Standard form
A number written as a × 10ᵏ, where a is at least 1 and less than 10 and k is an integer. Every positive number has exactly one such representation.
Significant figures
The digits of a number that carry information about its precision, counted from the first non-zero digit on the left.
Upper and lower bounds
For a value rounded to the nearest unit u, the true value lies in the half-unit interval from the rounded value − u/2 up to the rounded value + u/2.
Percentage error
ε = |vₐ − vₑ|/|vₑ| × 100, where vₐ is the approximate value used and vₑ is the exact or accepted value.
Logarithm
The exponent that answers the question "what power of b gives a?". Since a = bˣ is the same statement as x = log_b a, a logarithm is simply an exponent read backwards.
Explanation
Standard form. A number in standard form is written a × 10ᵏ with a at least 1 and less than 10, and k a whole number. The coefficient 41.7 × 10⁻⁴ is not in standard form because 41.7 is too large; rewriting it as 4.17 × 10⁻³ fixes both parts at once. To multiply, multiply the coefficients and add the exponents: (3 × 10⁴)(4 × 10⁻⁷) = 12 × 10⁻³ = 1.2 × 10⁻². To divide, divide the coefficients and subtract the exponents. To add or subtract, first write both numbers with the same power of ten, then add the coefficients. On a graphic display calculator the EE or ×10ˣ key enters the power directly, and the display mode can be set to SCI so that every answer comes back in standard form.
Laws of exponents. With the same base, aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ and (aᵐ)ⁿ = aᵐⁿ. Two consequences do most of the work in this course: a⁰ = 1 for any non-zero a, and a⁻ⁿ = 1/aⁿ, so a negative exponent means a reciprocal rather than a negative number. These laws are what make standard-form arithmetic work, because 10ᵐ × 10ⁿ = 10ᵐ⁺ⁿ.
Logarithms. The statement a = bˣ and the statement x = log_b a say exactly the same thing. In this course you only need two bases: base 10, written log x, and base e, written ln x. Both have their own calculator key. So 10^2.3 = 199.526… means log 199.526… = 2.3, and e^1.6 = 4.953… means ln 4.953… = 1.6. Logarithms appear whenever an unknown sits in the exponent, for example when you solve 1.08ⁿ = 2 for the doubling time by writing n = log 2 ÷ log 1.08 = 9.01. Note that only positive numbers have logarithms, because a positive base raised to any power is positive.
Rounding, bounds and accuracy. Unless a question says otherwise, give answers either exactly or to three significant figures. Significant figures are counted from the first non-zero digit, so 0.0298472 to 3 s.f. is 0.0298 while 0.0298472 to 3 d.p. is 0.030. A rounded value always hides an interval: a mass recorded as 24 g to the nearest gram lies between 23.5 g and 24.5 g, and these half-unit endpoints are the lower bound and the upper bound. Bounds propagate through a calculation: the largest possible area of a rectangle measuring 12.4 cm by 7.8 cm, each to 1 d.p., is 12.45 × 7.85 = 97.7325 cm².
Percentage error and estimation. Percentage error compares an approximate value with an accepted one: ε = |vₐ − vₑ|/|vₑ| × 100. The modulus signs mean the answer is never negative, and the exact value goes on the bottom. Using 3.14 in place of π gives ε = |3.14 − π|/π × 100 = 0.0507%, small enough for most practical work; a measurement 3% out may or may not matter depending on what it is used for, so always comment in context. A separate check is estimation: round every quantity to one significant figure and work the calculation out mentally. For (48.6 × 197) ÷ 1.04 the estimate is (50 × 200) ÷ 1 = 10 000, which sits close to the calculator value 9206, so no key has been mistyped.
Rounding too early. Rounding at an intermediate stage feeds a slightly wrong number into every later step, and the error grows. If a side length is stored as 8.246 but written down as 8.2, cubing it gives 551.4 instead of 560.7, an error of 1.7% created entirely by the student. The discipline is simple: keep full calculator accuracy in the machine, using ANS or a stored variable, and round once, at the very end.
Formulae
a × 10ᵏ with a at least 1 and less than 10, and k an integer
aᵐ × aⁿ = aᵐ⁺ⁿ; aᵐ ÷ aⁿ = aᵐ⁻ⁿ; (aᵐ)ⁿ = aᵐⁿ; a⁰ = 1; a⁻ⁿ = 1/aⁿ
a = bˣ ⟺ x = log_b a, for a greater than 0 and b greater than 0 with b ≠ 1
Percentage error ε = |vₐ − vₑ|/|vₑ| × 100
Bounds of a value rounded to the nearest u: rounded value ± u/2
Worked examples
q: A laboratory records the mass of a sample as 4.7 × 10⁻³ kg. The accepted mass is 4.52 × 10⁻³ kg. (a) Calculate the percentage error in the recorded mass. (b) The mass 4.7 × 10⁻³ kg was rounded to 2 significant figures. State its upper and lower bounds. (c) Comment on the reliability of the recorded value.
steps: (a) Identify the values: vₐ = 4.7 × 10⁻³ is the approximate (recorded) value and vₑ = 4.52 × 10⁻³ is the exact (accepted) value, which goes in the denominator.
(a) |vₐ − vₑ| = |4.7 × 10⁻³ − 4.52 × 10⁻³| = 0.18 × 10⁻³ = 1.8 × 10⁻⁴ kg.
(a) ε = (1.8 × 10⁻⁴) ÷ (4.52 × 10⁻³) × 100 = 3.9823… = 3.98% to 3 significant figures.
(b) Rounding to 2 s.f. here means rounding to the nearest 0.1 × 10⁻³ kg, so the half-unit is 0.05 × 10⁻³. Lower bound 4.65 × 10⁻³ kg, upper bound 4.75 × 10⁻³ kg.
(c) The accepted value 4.52 × 10⁻³ kg lies outside those bounds, so the difference is not explained by rounding alone: there is a systematic problem with the measurement, and an error of about 4% would be unacceptable for analytical work.
In context
Engineers quote a tolerance such as 250 mm ± 0.5 mm, which is exactly a statement of upper and lower bounds.
Scientific data are almost always given in standard form, so that 6.02 × 10²³ and 1.6 × 10⁻¹⁹ can sit in the same table without pages of zeros.
Common misconceptions
Students often write 0.45 × 10⁶ or 45 × 10⁴ and call it standard form. In fact the coefficient must be at least 1 and less than 10, so the correct form is 4.5 × 10⁵.
Students often put the approximate value on the bottom of the percentage error fraction. In fact the exact or accepted value vₑ is always the denominator, because the error is being measured as a fraction of the true size.
Students often confuse decimal places with significant figures. In fact 0.00846 has 3 significant figures but 5 decimal places, and the two instructions give different answers.
Students often round part (a) to 3 s.f. and then use that rounded number in part (b). In fact the unrounded value should stay in the calculator, because rounding early shifts every later answer.
Study tips
Set your calculator to display several decimal places while you work, and round only the final line of each part.
When a question says "give your answer in the form a × 10ᵏ", check the coefficient really is between 1 and 10 before writing it down; a single misplaced power of ten costs the accuracy mark.
Write the percentage error formula with the numbers substituted before evaluating it — that alone can earn method marks even if the arithmetic slips.
Summary
Standard form is a × 10ᵏ with a at least 1 and less than 10; multiply coefficients and add exponents.
a = bˣ and x = log_b a are the same statement; use log for base 10 and ln for base e.
Round to 3 significant figures unless told otherwise, and give bounds as the half-unit interval around a rounded value.
Percentage error is |vₐ − vₑ|/|vₑ| × 100, with the accepted value on the bottom; estimate with 1 s.f. values to sanity-check the calculator.
Check your understanding
Question 1: Written in the form a × 10ᵏ, the number 0.00417 is…
1. 4.17 × 10³
2. 4.17 × 10⁻³
3. 41.7 × 10⁻⁴
Question 2: Rounded to 3 significant figures, 0.0298472 is…
1. 0.0298
2. 0.030
3. 0.0299
Question 3: A length is given as 24 cm to the nearest centimetre. Its upper bound is…
1. 24.9 cm
2. 25 cm
3. 24.5 cm