Logarithmic and Exponential Functions

• Say it once and it stays: “A logarithm is the exponent.” When a question looks unfamiliar, translate every logarithm back into the exponent it stands for and the unfamiliarity usually disappears.
• Before you write anything in this topic, run S-C-O-P-E: State the conditions, Connect the representations, Operate one valid step at a time, Preserve exactness, Evaluate by substituting back.
• Two questions before any manipulation: is the base valid? and is every argument positive? Written at the top of the working, they cost one line and stop an invalid root from reaching your final answer.
• If you can do only one thing before the examination, do this: take the six “danger zone” statements and write, from memory, one sentence each on why the wrong version is wrong. Everything in this chapter hangs off those six reasons.

• State the conditions \(a>0\) and \(a\ne1\) that make \(\log_a x\) and \(a^{x}\) meaningful, and explain why each one is needed.
• Convert freely between \(y=a^{x}\) and \(x=\log_a y\), and use the notations \(\ln x=\log_e x\) and \(\lg x=\log_{10}x\) correctly.
• Explain that \(f(x)=e^{x}\) and \(g(x)=\ln x\) are inverse functions, and use \(\ln(e^{x})=x\) and \(e^{\ln x}=x\) with their correct conditions.
• State the domain, range, intercept, asymptote and increasing or decreasing behaviour of \(e^{x}\) and of \(\ln x\).
• Sketch \(y=e^{x}\), \(y=\ln x\) and \(y=x\) on one pair of axes and show the reflection, including how a point \((a,b)\) maps to \((b,a)\).
• Compare \(a^{x}\) for \(a>1\) with \(a^{x}\) for \(0<a<1\), and match each to its logarithmic inverse.
• Sketch \(y=ke^{nx}+a\), stating the horizontal asymptote \(y=a\) and describing every transformation from \(y=e^{x}\) in the correct order.
• Sketch \(y=k\ln(ax+b)\), finding the domain from \(ax+b>0\) and the vertical asymptote from \(ax+b=0\).
• State and apply the product, quotient and power laws, with the positivity conditions attached rather than assumed.
• Combine an expression such as \(3+2\lg p-\lg q\) into a single logarithm, and expand a single logarithm back into separate terms.
• Explain, with a counterexample, why \(\log(M+N)\) cannot be split.
• State and use the change-of-base law \(\log_a b=\dfrac{\log_c b}{\log_c a}\), both to obtain an exact value and to obtain a calculator value.
• Solve a logarithmic equation and reject any root that would make an argument zero or negative.
• Solve \(a^{x}=b\) for \(a>0\), \(a\ne1\), \(b>0\), giving \(x=\dfrac{\ln b}{\ln a}\).
• Decide, before starting, whether the bases can be matched exactly, and match them when they can.
• Take logarithms of both sides correctly when the bases cannot be matched, including when the exponent is a linear expression.
• Leave answers in exact logarithmic form, and give a decimal only when the question asks for one.
• Verify a solution by substituting it back into the original equation.

Why this topic exists. Any quantity that multiplies itself over equal steps — compound growth, radioactive decay, cooling, sound intensity, the pH scale — is naturally described by an exponential function, and the only way to get the unknown out of the exponent is a logarithm. Later in this course the same two functions reappear as the ones whose derivatives and integrals you are expected to know, so a shaky grasp here becomes a shaky grasp in calculus.

• Letting a logarithm argument be zero: writing \(\log_a 0\). Why it fails \(\log_a 0\) would be the exponent \(x\) with \(a^{x}=0\). For \(a>0\), \(a^{x}\) is positive for every real \(x\) and has no smallest value, so no exponent produces zero. Do this Treat any candidate that makes an argument zero as outside the domain and reject it.
• Letting a logarithm argument be negative: writing \(\log_2(-8)=-3\) because \(2^{-3}\) “looks negative”. Why it fails \(2^{-3}=\tfrac18\), which is positive. A negative exponent gives a small positive number, never a negative one, so \(2^{x}\) can never equal \(-8\). Do this Separate the two ideas: the value of a logarithm may be negative; its argument may not.
• Using base \(1\), or a negative base, for a real logarithm. Why it fails \(1^{x}=1\) for every \(x\), so \(\log_1\) cannot single out an exponent. A negative base fails earlier still: \((-4)^{1/2}\) is not real, so \(a^{x}\) is not even defined across the reals. Do this Check \(a>0\) and \(a\ne1\) whenever a base is unknown or is being found.
• Treating \(\ln x\) as defined for all real \(x\), for instance evaluating \(\ln(-5)\) or building a table of values that starts at \(x=0\). Why it fails \(\ln\) is the inverse of \(e^{x}\), whose range is \(y>0\). An inverse can only accept what the original produced, so the domain of \(\ln\) is \(x>0\). Do this Before any table, sketch or substitution, write down the domain condition first.
• Splitting a sum: \(\log(M+N)=\log M+\log N\). Why it fails The laws mirror index laws, and there is no index law for \(a^{p}+a^{q}\) — a sum of powers does not simplify. One substitution settles it: \(\lg(1+99)=\lg 100=2\), but \(\lg 1+\lg 99=0+1.9956\ldots\) Do this Leave a sum inside the logarithm. If a question seems to need it split, look for a factorisation instead.
• Reversing the quotient law: \(\log_a\!\left(\dfrac MN\right)=\log_a N-\log_a M\). Why it fails It comes from \(\dfrac{a^{p}}{a^{q}}=a^{p-q}\), and the numerator’s exponent is the one that survives first. Test: \(\lg\!\left(\tfrac{100}{10}\right)=1\), and \(\lg 100-\lg 10=1\), while \(\lg 10-\lg 100=-1\). Do this Say “top minus bottom” out loud as you write it.
• Dropping the multiplier from the power law: writing \(\log_a\!\left(M^{3}\right)=\log_a M\). Why it fails \(M^{3}=M\times M\times M\), so by the product law its logarithm is \(\log_a M\) three times over. The \(3\) is not decoration; it is a count. Do this Check with numbers: \(\lg\!\left(10^{3}\right)=3\), not \(1\).
• Losing the positivity conditions once the logarithms have been combined into one. Why it fails Combining widens the set of values that make the expression legal. In \(\log_2 x+\log_2(x-2)\) both \(x\) and \(x-2\) must be positive, but the combined \(\log_2\!\left(x(x-2)\right)\) is also legal at \(x=-2\), where the original is not. The extra root is manufactured by the combining step. Do this Write the conditions on the first line and test every candidate against the original equation.
• Moving a vertical translation into the exponent: reading \(3e^{-2x}+4\) as \(3e^{-2x+4}\). Why it fails The \(+4\) is applied after the exponential has been evaluated, so it raises every point — and the asymptote — by \(4\). Inside the exponent it would instead multiply the curve by the constant \(e^{4}\): \(3e^{-2x+4}=3e^{4}e^{-2x}\), whose asymptote is still \(y=0\) and whose \(y\)-intercept is \(3e^{4}\approx163.8\), not \(7\). Do this Evaluate at \(x=0\) as a check: \(3e^{0}+4=7\) tells you immediately which reading you have.
• Confusing horizontal and vertical asymptotes: giving \(x=4\) as the asymptote of \(y=3e^{-2x}+4\), or \(y=2\) as the asymptote of \(y=2\ln(3x-6)\). Why it fails An exponential is bounded in the \(y\)-direction and unbounded in \(x\), so its asymptote must be horizontal. A logarithm is the reverse. The directions are exchanged precisely because the functions are inverses. Do this Ask which variable is restricted. Restricted \(y\) means a horizontal asymptote; restricted \(x\) means a vertical one.
• Saying an exponential graph “reaches” or “touches” its asymptote. Why it fails It would contradict the range. \(e^{x}>0\) for every real \(x\); there is no value of \(x\) at which \(e^{x}=0\), however large and negative \(x\) becomes. \(e^{-100}\approx3.7\times10^{-44}\), and still not zero. Do this Use “approaches”, “tends to” or “gets arbitrarily close to”, and never draw the curve meeting the line.
• Treating \(e^{x}\) and \(\ln x\) as unrelated functions that happen to appear in the same chapter. Why it fails It throws away the most useful fact available. Because they are inverses, \(\ln\) is the tool that removes \(e\), and \(e\) is the tool that removes \(\ln\). Students who miss this try to solve \(e^{3x}=7\) by dividing. Do this When an unknown is stuck inside one of them, apply the other to both sides.
• Replacing an exact answer with a decimal when exactness was required: writing \(x=1.39\) instead of \(\tfrac12\!\left(1+\dfrac{\ln 7}{\ln 3}\right)\). Why it fails The decimal is an approximation to the answer, not the answer, and on a non-calculator paper it cannot be obtained at all. Exact answers in this course may legitimately contain \(e\), \(\ln\), surds and fractions. Do this Give the exact form. Add a decimal only if the question asks for one, and then state the accuracy.
• Rounding logarithms during the intermediate working, for instance using \(\ln 3=1.10\) and then continuing. Why it fails The error propagates and grows. \(\dfrac{\ln 7}{\ln 3}=1.7712\ldots\), but \(\dfrac{1.95}{1.10}=1.7727\ldots\), already wrong in the third decimal place. That particular slip happens to survive rounding to 3 significant figures; round a little harder and it does not, since \(\dfrac{1.9}{1.1}=1.7273\ldots\) gives \(1.73\) where the true value gives \(1.77\). You cannot tell in advance which case you are in, which is why the rule is to round once, at the end. Do this Carry the symbols \(\ln 7\) and \(\ln 3\) to the last line, then evaluate once.
• Taking logarithms of a side that is not positive, for instance “solving” \(2^{x}=-5\) as \(x=\dfrac{\ln(-5)}{\ln 2}\). Why it fails \(\ln(-5)\) does not exist, so the expression written down is meaningless rather than merely unhelpful. Applying \(\ln\) to both sides is only valid when both sides are strictly positive. Do this Check \(b>0\) first. If it fails, the complete answer is “no real solution”, with a reason.
• Forgetting to divide by the logarithm of the base: going from \(x\ln 2=\ln 5\) to \(x=\ln 5-\ln 2\). Why it fails \(\ln 2\) multiplies \(x\), so it is removed by division, not by subtraction. Numerically \(\dfrac{\ln 5}{\ln 2}=2.3219\ldots\) while \(\ln 5-\ln 2=0.9163\ldots\); only the first satisfies \(2^{x}=5\). Do this Treat \(\ln 2\) as the ordinary number \(0.693\ldots\) and ask what you would do to remove it.
• Describing \(y=e^{-2x}\) as “a reflection of \(y=e^{x}\) in the \(y\)-axis” and stopping there. Why it fails Replacing \(x\) by \(-2x\) does two things: the minus sign reflects, and the \(2\) compresses horizontally by scale factor \(\tfrac12\). The reflection alone would give \(e^{-x}\), a visibly slower curve: at \(x=1\), \(e^{-1}=0.368\) but \(e^{-2}=0.135\). Do this Name both transformations, and say which scale factor applies to which direction.
• Introducing series expansions of \(e^{x}\) or \(\ln(1+x)\) as though they were required content. Why it fails They are not in this syllabus, so they cannot be assumed, cannot be quoted as justification, and consume time that the marked method needs. A “show that” answered by series is not answering the question asked. Do this Stay with the definitions, the graphs, the four laws and the equation-solving methods in this chapter. They cover every Topic 6 requirement.

• Every candidate takes both 0606 papers. The qualification is untiered, so neither paper is reserved for one group of candidates and there is no tiered alternative to either. Each paper is two hours and 80 marks, each carries half the qualification, all questions are compulsory, and necessary working must be shown. Either paper may draw on any part of the subject content, so Topic 6 can appear in either. What changes between them is not the mathematics but what you are allowed to press. The examination is available in the June and November series, and in the March series in India, for this cycle, and the candidate grade range is A* to E.
• The formula sheet does not help you here. The supplied list covers the circle equation, mensuration, the quadratic formula, the binomial theorem, progressions, three trigonometric identities, and the sine rule, cosine rule and area of a non-right-angled triangle. The laws of logarithms are not on it. Product, quotient, power and change of base have to come out of your own memory in both papers.
• The accuracy contract for this course applies with unusual force to logarithms, because rounding early destroys the very thing being tested.
• Give answers in simplest form unless told otherwise. \(\ln 8\) and \(3\ln 2\) are both acceptable; \(\ln 2 + \ln 2 + \ln 2\) is not simplest.
• Exact answers may legitimately contain \(e\), \(\ln\) and fractions. Do not treat a logarithm as something that must be evaluated.
• If a non-exact answer is wanted, give at least 3 significant figures, and keep extra accuracy in the intermediate working.