P2.5 Exponentials and logarithms

What does the graph of y = aˣ look like?

In P1 the variable was the base, as in \(x^2\) or \(x^3\). Now the base is fixed and the variable is the power. That small change produces a completely different kind of curve: one that grows by the same factor every time \(x\) goes up by 1.

Why must \(a\) be positive and not 1? If \(a = 1\) the graph is just the line \(y = 1\), which is not interesting. If \(a\) were negative, numbers like \((-2)^{1/2}\) would not be real, so the curve would have gaps everywhere. With \(a > 0\), the power \(a^x\) is defined for every real \(x\).

Four features follow from the index laws:

• Intercept. \(a^0 = 1\) for every base, so every curve \(y = a^x\) passes through \((0, 1)\). At \(x = 1\) the height is \(a\) itself.
• Always positive. \(a^x > 0\) for every real \(x\). The curve never meets the \(x\)-axis, which is a horizontal asymptote.
• Growth or decay. If \(a > 1\) the curve increases, more and more steeply (exponential growth). If \(0 < a < 1\) it decreases towards the \(x\)-axis (exponential decay).
• Reflection. \(\left(\tfrac{1}{2}\right)^x = 2^{-x}\), so \(y = \left(\tfrac12\right)^x\) is the reflection of \(y = 2^x\) in the \(y\)-axis.

The P1 transformations still apply. \(y = a^x + k\) moves the curve and its asymptote up by \(k\), so the asymptote becomes \(y = k\). \(y = a^{x+c}\) moves the curve \(c\) to the left; the asymptote stays on the \(x\)-axis but the \(y\)-intercept becomes \(a^c\).

What is a logarithm?

Suppose \(2^x = 8\). You can see that \(x = 3\). But for \(2^x = 7\) there is no whole-number answer, and we need a name for "the power of 2 that gives 7". That name is the logarithm.

Moving between the two forms is the key skill of the whole topic. For example \(10^3 = 1000 \iff \log_{10} 1000 = 3\), and \(2^{-3} = \frac18 \iff \log_2 \frac18 = -3\). Notice that a logarithm can be negative or a fraction; only the number you take the log of has to be positive.

• Two special values: \(\log_a a = 1\) because \(a^1 = a\), and \(\log_a 1 = 0\) because \(a^0 = 1\).
• Since \(a^x\) is always positive, \(\log_a b\) exists only for \(b > 0\). There is no log of 0 or of a negative number.
• \(y = \log_a x\) undoes \(y = a^x\): it is the inverse function. So \(\log_a a^x = x\) and \(a^{\log_a x} = x\).

Because the two functions are inverses, their graphs are reflections of each other in the line \(y = x\). Every point \((p, q)\) on \(y = 2^x\) becomes the point \((q, p)\) on \(y = \log_2 x\). The intercept \((0, 1)\) becomes \((1, 0)\), and the horizontal asymptote \(y = 0\) becomes the vertical asymptote \(x = 0\).

Every log curve with base greater than 1 has the same shape: through \((1, 0)\), increasing, slowly. A larger base gives a flatter curve, because a larger base needs a smaller power to reach the same number.

What are the laws of logarithms?

Each law of logarithms is an index law read backwards. Suppose \(\log_a x = m\) and \(\log_a y = n\), so \(x = a^m\) and \(y = a^n\). Then \(xy = a^m a^n = a^{m+n}\), so \(\log_a xy = m + n = \log_a x + \log_a y\). Multiplying numbers adds their powers, so the log of a product is the sum of the logs. The other laws come the same way from \(a^m \div a^n = a^{m-n}\) and \((a^m)^k = a^{mk}\).

The reciprocal law is just the power law with \(k = -1\). Roots are powers too: \(\log_a \sqrt{x} = \frac12 \log_a x\). There is no law for \(\log(x + y)\), and \(\log x \div \log y\) is not \(\log \frac{x}{y}\).

How do I solve aˣ = b?

If both sides are obviously powers of the same base, match the powers, as in \(3^x = 81 = 3^4\). Otherwise, take logs of both sides (any base will do; \(\log_{10}\) is the calculator key) and use the power law to bring the unknown down from the power:

This result is not in the formulae booklet; derive it by taking logs each time. A solution exists only if \(b > 0\), because \(a^x\) is never zero or negative. Graphically, the solution of \(a^x = b\) is the \(x\)-coordinate where the curve \(y = a^x\) meets the horizontal line \(y = b\); since an exponential curve is always increasing (or always decreasing), there is exactly one such point.

When the unknown is in two powers with different bases, take logs, expand the brackets, collect the \(x\) terms on one side and factorise. Keep the exact form (such as \(\frac{\log 40}{\log 5}\)) until the last line and round only the final answer.

How do I solve a quadratic in aˣ?

Look for \(a^{2x}\) together with \(a^x\). Because \(a^{2x} = (a^x)^2\), the substitution \(u = a^x\) turns the equation into an ordinary quadratic in \(u\). Before substituting, rewrite every term using the index laws:

• \(4^x = (2^2)^x = (2^x)^2\), so \(4^x = u^2\) when \(u = 2^x\).
• \(2^{x+3} = 2^3 \times 2^x = 8u\).
• \(3^{2x+1} = 3 \times (3^x)^2 = 3u^2\) when \(u = 3^x\).

Solve for \(u\), then solve \(a^x = u\) for each value. Reject any \(u \le 0\): \(a^x\) is never zero or negative, so such a value gives no \(x\) at all.

How do I change the base of a logarithm?

A calculator works in base 10 (and base e, met in P3). To evaluate a log in any other base, or to combine logs with different bases in one equation, rewrite them in a common base.

Proof. Let \(y = \log_a x\), so \(a^y = x\). Take \(\log_b\) of both sides: \(\log_b a^y = \log_b x\), so \(y\log_b a = \log_b x\), giving \(y = \frac{\log_b x}{\log_b a}\).

Two special cases appear often. Putting \(x = b\) gives \(\log_a b = \frac{1}{\log_b a}\). Putting \(a = b^2\) gives \(\log_{b^2} x = \frac12\log_b x\); for example \(\log_9 x = \frac12\log_3 x\).

How do I solve a logarithmic equation?

A logarithmic equation has the unknown inside one or more logs. The method has four steps, and the last one is the step most often forgotten.

1. Make every log the same base (change of base if needed).
2. Use the laws to collect the logs into a single log on one side, or one log on each side.
3. Remove the logs: \(\log_a P = k \Rightarrow P = a^k\), and \(\log_a P = \log_a Q \Rightarrow P = Q\). A plain number \(k\) can be rewritten as \(\log_a a^k\) if that helps.
4. Solve the resulting equation, then check every root in the original equation: each log must have a positive argument.

Why the check? Combining logs can enlarge the set of allowed values. \(\log_3 (x-1)^2\) is defined for every \(x \ne 1\), but the original \(2\log_3(x-1)\) needs \(x > 1\). The algebra therefore can produce a root that the original equation never allowed.

Exam tips for P2.5

Common misconceptions

How this topic connects

P1 Pure Mathematics 1

The laws of indices are the laws of logarithms written the other way round. Translations and reflections of graphs give the sketches of \(a^x + k\) and \(a^{-x}\), and factorising quadratics finishes every hidden-quadratic question.

P2 sequences and series

Finding the least \(n\) for which a geometric term or sum passes a value means solving \(r^n = k\) with a logarithm, with care over the inequality sign when \(\log r < 0\).

P3 Pure Mathematics 3

The base e, \(\ln x\) as the inverse of \(e^x\), exponential growth and decay models, and turning \(y = ax^n\) or \(y = kb^x\) into straight lines by taking logs all build on this page. In IAL Physics, radioactive decay, capacitor discharge and half-life use the same exponential functions.

Key formulae

Check your understanding

Specification checklist

• (P2 5.1) I can sketch \(y = a^x\) for \(a > 1\) and for \(0 < a < 1\), showing the point \((0, 1)\) and the asymptote.
• (P2 5.1) I can sketch simple transformations of \(y = a^x\), giving intercepts and the equation of the asymptote.
• (P2 5.2) I can switch between index form \(a^x = b\) and log form \(\log_a b = x\), and use \(\log_a a = 1\) and \(\log_a 1 = 0\).
• (P2 5.2) I can use the product, quotient, power and reciprocal laws of logarithms to combine or split logs, and express logs in terms of given letters.
• (P2 5.3) I can solve \(a^x = b\), including equations with the unknown in two powers, giving exact or rounded answers.
• (P2 5.3) I can solve equations that reduce to a quadratic in \(a^x\), rejecting impossible values with a reason.
• (P2 5.3) I can use the change of base formula, and solve simple logarithmic equations, checking that every log in the original equation is defined.