Cambridge IGCSE Additional Mathematics · Syllabus 0606 · Logarithmic and Exponential Functions
Logarithm
What is Logarithm?
The exponent to which a fixed base must be raised to produce a given positive number. For a base a with a greater than zero and a not equal to one, and for a positive number y, the statement log to base a of y equals x means exactly the same as a raised to the power x equals y, so a logarithm answers the question of what power of the base gives that number. The base must be positive and not equal to one, and the number inside the logarithm, called the argument, must be strictly positive, because no real power of a positive base can produce zero or a negative value.
This definition is part of the Logarithmic and Exponential Functions chapter in Cambridge IGCSE Additional Mathematics.
Logarithm in context
A logarithm is an exponent. Writing \(\log_a b\) asks one question and one question only: what power of \(a\) gives \(b\)? Everything in this chapter — the graphs, the asymptotes, the four laws, the equations — is a consequence of that single sentence and of the conditions that keep it meaningful.
Common mistakes with Logarithm
- 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.
- 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.
- 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.
- 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.
- 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.