Logarithmic and exponential functions

• A logarithm is an index. To free an unknown from an index, take ln of both sides; before you divide an inequality by \(\ln a\), ask whether \(a < 1\). On a log–linear graph the intercept is \(\ln k\), never \(k\).
• \(\log_a b = c \iff a^c = b\). Take ln of both sides, and check the sign of \(\ln(\text{base})\) before you divide. The intercept is \(\ln k\).

• 3.2.1 I can understand — understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base)
• 3.2.2 I can understand — understand the definition and properties of eˣ and ln x, including their relationship as inverse functions and their graphs
• 3.2.3 I can use — use logarithms to solve equations and inequalities in which the unknown appears in indices
• 3.2.4 I can use — use logarithms to transform a given relationship to linear form, and hence determine unknown constants by considering the gradient and/or intercept

Accuracy for this chapter. Leave exact answers exact: \(\ln 7\), \(e^2\), \(\dfrac{2}{e - 1}\), \(\dfrac{\ln 5}{\ln 2}\). Give a decimal to 3 significant figures only when the question asks for one, and show the exact quotient of logarithms before it; a value read straight off a calculator with no method line earns nothing. Tabulate logarithms to 4 decimal places. Keep unrounded values (1.484 43, 13.425, 3.9905) until the end, and keep an integer-inequality boundary unrounded until you have chosen the integer.

• “\(0.5^x < 0.1 \Rightarrow x\ln 0.5 < \ln 0.1 \Rightarrow x < \dfrac{\ln 0.1}{\ln 0.5} = 3.32\).” Correct \(\ln 0.5 = -0.693\) is negative, and dividing an inequality by a negative number reverses it: \(x > 3.32\). The logarithm of any base between 0 and 1 is negative. Write its sign on the line before you divide (section C).
• “The graph of \(\ln y\) against \(x\) has intercept 1.2, so \(k = 1.2\).” Correct The intercept is \(\ln k\), not \(k\): \(\ln k = 1.2\) gives \(k = e^{1.2} = 3.32\). In the same way, for \(y = k(a^x)\) the gradient is \(\ln a\), not \(a\) (section D).
• “\(\log(x + y) = \log x + \log y\).” Correct There is no law for the logarithm of a sum. The product law runs the other way: \(\log x + \log y = \log(xy)\).
• “\(3^{x + 1} = 4^{2x - 1}\), so \(x + 1 = 2x - 1\).” Correct Indices may be equated only when the bases are equal. Take ln of both sides and keep \(\ln 3\) and \(\ln 4\) as multipliers: \((x + 1)\ln 3 = (2x - 1)\ln 4\).
• “\(\log_2(x + 3) + \log_2(x - 1) = 5\) gives \(x = 5\) or \(x = -7\).” Correct \(x = -7\) makes \(x - 1 = -8\), and a negative number has no logarithm. Reject it, and write the reason. Only \(x = 5\) is a solution.
• “\(\ln x \to 0\) as \(x \to 0\).” Correct \(\ln x \to -\infty\) as \(x \to 0^{+}\): the \(y\)-axis is an asymptote, and \(\ln 0\) does not exist.
• “\(\log(x + y) = \log x + \log y\).” Repair There is no law for the logarithm of a sum. The product law is \(\log x + \log y = \log(xy)\). Test with numbers: \(\log_{10}(2 + 8) = 1\), but \(\log_{10} 2 + \log_{10} 8 = \log_{10} 16 = 1.20\).
• “\(\log_{10} 12 \div \log_{10} 4 = \log_{10} 3\).” Repair The quotient law is about a difference: \(\log_{10} 12 - \log_{10} 4 = \log_{10} 3\). A quotient of logarithms does not simplify by the laws: \(\dfrac{\log_{10} 12}{\log_{10} 4} = 1.79\), while \(\log_{10} 3 = 0.477\).
• “\((\log_{10} x)^2 = 2\log_{10} x\).” Repair \(2\log_{10} x = \log_{10} x^2\). \((\log_{10} x)^2\) is the square of the number \(\log_{10} x\), and an equation containing it is a quadratic in \(u = \log_{10} x\).
• Accepting \(x = -7\) as a solution of \(\log_2(x + 3) + \log_2(x - 1) = 5\). Repair \(x - 1 = -8\) has no logarithm. Reject it and say why; only \(x = 5\) survives. Test every root in the original equation, not in the rearranged quadratic.
• “\(2^x = 5 \Rightarrow x = \tfrac{5}{2}\)”, or “\(x = \sqrt{5}\)”. Repair The unknown is in the index, so neither division nor a root frees it. Take ln: \(x\ln 2 = \ln 5\), so \(x = \dfrac{\ln 5}{\ln 2} = 2.32\). Check: \(2^{2.32} = 4.99\).
• “\(3^{x + 1} = 4^{2x - 1} \Rightarrow x + 1 = 2x - 1\).” Repair Indices can be equated only when the bases are equal. Take ln of both sides and keep \(\ln 3\) and \(\ln 4\) as multipliers: \((x + 1)\ln 3 = (2x - 1)\ln 4\), giving \(x = 1.48\).
• “\(\dfrac{\ln 5}{\ln 2} = \ln 2.5\).” Repair \(\ln 2.5 = \ln 5 - \ln 2 = 0.916\), a difference. \(\dfrac{\ln 5}{\ln 2}\) is a quotient of two numbers, \(2.32\). Leave it as a quotient in the exact answer.
• “\(0.5^x < 0.1 \Rightarrow x < \dfrac{\ln 0.1}{\ln 0.5} = 3.32\).” Repair \(\ln 0.5\) is negative, so dividing reverses the inequality: \(x > 3.32\). Check with \(x = 3\) (\(0.125\), fails) and \(x = 4\) (\(0.0625\), works).
• “\(0.8^n < 0.05\) gives \(n > 13.4\), so the smallest \(n\) is 13.” Repair 13 is less than 13.4, so it does not satisfy \(n > 13.4\): the smallest integer greater than the boundary 13.425 is 14. Go up to the next integer, then check both neighbours: \(0.8^{13} = 0.0550\) fails, \(0.8^{14} = 0.0440\) works.
• “\(\ln x \to 0\) as \(x \to 0\).” Repair \(\ln x \to -\infty\) as \(x \to 0^{+}\); the \(y\)-axis is an asymptote, and \(\ln 0\) does not exist.
• Reading the intercept of a \(\ln y\) against \(x\) graph as \(k\). Repair The intercept is \(\ln k\), so \(k = e^{\text{intercept}}\). An intercept of 1.2 means \(k = 3.32\).
• Reading the gradient of a \(\ln y\) against \(x\) graph as \(a\) in \(y = k(a^x)\). Repair The gradient is \(\ln a\), so \(a = e^{\text{gradient}}\). A gradient of 0.405 means \(a = 1.50\).
• For \(y = kx^n\), plotting \(\ln y\) against \(x\). Repair A power law is linear in \(\ln x\) and \(\ln y\); plotting \(\ln y\) against \(x\) gives a curve. Only the exponential law \(y = k(a^x)\) is linear in \(x\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• What the laws do not say. There is no law for \(\log(x + y)\): it does not split. And a quotient of logarithms is not the logarithm of a quotient: \(\dfrac{\log_{10} 12}{\log_{10} 4} = 1.79\), but \(\log_{10} 3 = 0.477\). Also keep \((\log x)^2\), the square of the number \(\log x\), apart from \(\log x^2 = 2\log x\).
• Two traps are built into this outcome. The intercept is \(\ln k\), not \(k\): a line through \((0, 1.2)\) means \(k = e^{1.2} = 3.32\), not 1.2. And for the exponential law the gradient is \(\ln a\), not \(a\): a gradient of 0.405 means \(a = e^{0.405} = 1.50\).
• Interleave with the chapters that use this one. When you reach chapter 12, re-sketch \(e^x\) and \(\ln x\) before you meet their derivatives; in chapter 13, re-state \(\ln x = y \iff e^y = x\) and the laws before you use them in integration; in chapter 14, sketch \(y = e^x\) and \(y = 3 - x\) to see why \(e^x = 3 - x\) has one root; and in chapter 16, redo worked example 7, because finding when a solution of the form \(y = Ae^{bx}\) reaches a given value takes the same last step: take ln of both sides. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: an individual examination question may involve ideas and methods from more than one section of that paper’s content, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3 (and Paper 2 on the AS-only pure route). Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. The Paper 2 (Pure Mathematics 2) outcomes this chapter also serves are marked in the syllabus map: Paper 2 is offered only in the AS-only Pure Mathematics route (Papers 1 and 2), where it is 40% of the AS Level, and that route cannot be carried forward to the A Level. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• The syllabus outcomes lend themselves to four question forms: a logarithmic equation to be solved, where rejecting a root with its reason is part of the answer; an index equation or inequality solved by taking logarithms, sometimes asking for the smallest integer; a “show that” that turns an equation into a quadratic in \(e^x\), followed by “hence solve”; and a straight-line graph of \(\ln y\) against \(x\) or \(\ln x\), from which the constants are found and then used. A sketch of a transformed \(e^x\) or \(\ln x\) can open a longer question that then solves an equation.
• MF19 prints none of this section. The definition \(\log_a b = c \iff a^c = b\), the three laws, \(\ln x = y \iff e^y = x\), the graphs of \(e^x\) and \(\ln x\), and both linear forms must be known. The MF19 card collects them.
• Write the line \((x + 1)\ln 3 = (2x - 1)\ln 4\) before any number: a value read off a calculator with no method line earns nothing. Give the exact form (\(\ln 7\), \(\dfrac{\ln 5}{\ln 2}\), \(\dfrac{2}{e - 1}\)) before the decimal, round once to 3 significant figures, write the sign of \(\ln(\text{base})\) before dividing an inequality, and keep an integer boundary such as 13.425 unrounded until the integer is chosen.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.