The function y = aˣ for a > 0, a ≠1: its graph, intercept, asymptote and translations

The meaning of logₐ x, including logₐ a = 1 and logₐ 1 = 0

Why y = logₐ x is the reflection of y = aˣ in the line y = x

The laws of logarithms for products, quotients, powers and reciprocals

Solving aˣ = b, including equations with the unknown in two different powers

Equations that become quadratics in aˣ

Changing the base of a logarithm, and proving the change of base formula

Solving logarithmic equations and checking that every log is defined

1. A logarithm answers the question "what power?": logₐ b = x means exactly the same as aˣ = b, and every exponential or logarithmic equation on this page is solved by moving between those two forms.

2. y = aˣ (a > 0, a ≠1) passes through (0, 1), is always positive and has the x-axis as an asymptote. It increases if a > 1 and decreases if 0 < a < 1.

3. logₐ a = 1, logₐ 1 = 0, and you can only take the log of a positive number.

4. Laws: log xy = log x + log y; log x/y = log x - log y; log xᵏ = klog x; log 1/x = -log x. All logs in one law must have the same base.

5. To solve aˣ = b, take logs of both sides: x = log b/log a. If a²ˣ and aˣ both appear, let u = aˣ, solve a quadratic and reject u ≤0.

6. Change of base: logₐ x = (log_b x)/(log_b a). For log equations, combine into one log, remove the logs, and check every root keeps each original argument positive.

In P1 the variable was the base, as in x² or x³. 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 aˣ is defined for every real x.

Four features follow from the index laws:

Intercept. a⁰ = 1 for every base, so every curve y = aˣ passes through (0, 1). At x = 1 the height is a itself.

Always positive. aˣ > 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. (½)ˣ = 2⁻ˣ, so y = (½)ˣ is the reflection of y = 2ˣ in the y-axis.

The P1 transformations still apply. y = aˣ + k moves the curve and its asymptote up by k, so the asymptote becomes y = k. y = aˣ⁺ᶜ moves the curve c to the left; the asymptote stays on the x-axis but the y-intercept becomes aᶜ.

WORKED EXAMPLE · QUESTION

The curve C has equation y = 3ˣ - 9. (a) Find the coordinates of the points where C meets the coordinate axes. (b) Write down the equation of the asymptote to C. (c) Sketch C. (d) Find the value of x for which y = 72.

SOLUTION

(a) On the y-axis, x = 0: y = 3⁰ - 9 = 1 - 9 = -8, giving (0, -8). On the x-axis, y = 0: 3ˣ = 9 = 3², so x = 2, giving (2, 0). (0, -8) and (2, 0)

(b) As x →-∞, 3ˣ →0, so y →-9. y = -9

(c) The curve y = 3ˣ moved down 9: increasing, through both intercepts, approaching y = -9 on the left (see the figure).

(d) 3ˣ - 9 = 72 ⇒3ˣ = 81 = 3⁴. x = 4

Suppose 2ˣ = 8. You can see that x = 3. But for 2ˣ = 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 10³ = 1000 ⇔log₁₀ 1000 = 3, and 2⁻³ = ⅛ ⇔log₂ ⅛ = -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: logₐ a = 1 because a¹ = a, and logₐ 1 = 0 because a⁰ = 1.

Since aˣ is always positive, logₐ b exists only for b > 0. There is no log of 0 or of a negative number.

y = logₐ x undoes y = aˣ: it is the inverse function. So logₐ aˣ = x and a^(logₐx) = 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 = 2ˣ becomes the point (q, p) on y = log₂ 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.

WORKED EXAMPLE · QUESTION

(a) Without a calculator, find the value of: (i) log₂ 64 (ii) log₃ (1/27) (iii) log₂₅ 5 (iv) log₄ 8 (v) log₇ 1. (b) Solve (i) logₓ 49 = 2, where x > 0, and (ii) log₅ y = -2.

SOLUTION

(a) Ask "what power?" each time. (i) 2⁶ = 64, so 6. (ii) 3⁻³ = (1/27), so -3. (iii) 25^(1/2) = 5, so ½. (iv) 4ˣ = 8 ⇒2²ˣ = 2³ ⇒x = (3/2). (v) 7⁰ = 1, so 0. 6, -3, ½, (3/2), 0

(b) Rewrite in index form. (i) x² = 49, and a base must be positive, so x = 7. (ii) y = 5⁻² = (1/25). x = 7, y = (1/25)

Each law of logarithms is an index law read backwards. Suppose logₐ x = m and logₐ y = n, so x = aᵐ and y = aⁿ. Then xy = aᵐ aⁿ = aᵐ⁺ⁿ, so logₐ xy = m + n = logₐ x + logₐ 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 aᵐ ÷aⁿ = aᵐ⁻ⁿ and (aᵐ)ᵏ = aᵐᵏ.

The reciprocal law is just the power law with k = -1. Roots are powers too: logₐ √x = ½ logₐ x. There is no law for log(x + y), and log x ÷log y is not log x/y.

WORKED EXAMPLE · QUESTION

(a) Show that 2log₃ 6 - log₃ 4 = 2. (b) Given that logₐ 2 = p and logₐ 3 = q, express in terms of p and q: (i) logₐ 18 (ii) logₐ(8/9a) (iii) logₐ(⅙).

SOLUTION

(a) Power law first, then the quotient law: 2log₃ 6 - log₃ 4 = log₃ 36 - log₃ 4 = log₃ (36/4) = log₃ 9 = 2, since 3² = 9. = 2 as required

(b)(i) 18 = 2 ×3², so logₐ 18 = logₐ 2 + 2logₐ 3. p + 2q

(b)(ii) Split the quotient and the product, and use logₐ a = 1: logₐ 2³ - logₐ 3² - logₐ a. 3p - 2q - 1

(b)(iii) Reciprocal law: -logₐ 6 = -(logₐ 2 + logₐ 3). -p - q

If both sides are obviously powers of the same base, match the powers, as in 3ˣ = 81 = 3⁴. Otherwise, take logs of both sides (any base will do; log₁₀ 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 aˣ is never zero or negative. Graphically, the solution of aˣ = b is the x-coordinate where the curve y = aˣ 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 log 40/log 5) until the last line and round only the final answer.

WORKED EXAMPLE · QUESTION

Solve, giving each answer to 3 significant figures: (a) 5ˣ = 40 (b) 2ˣ⁺¹ = 3ˣ.

SOLUTION

(a) Take logs of both sides: xlog 5 = log 40, so x = log 40/log 5 = 2.2920… x = 2.29

(b) Take logs and use the power law on each side: (x+1)log 2 = xlog 3. Expand and collect the x terms: xlog 2 + log 2 = xlog 3 ⇒log 2 = x(log 3 - log 2). Divide, and simplify with the quotient law:

x = 1.71

Check: 2^(2.7095) ≈3^(1.7095) ≈6.54, the crossing point in the figure.

Look for a²ˣ together with aˣ. Because a²ˣ = (aˣ)², the substitution u = aˣ turns the equation into an ordinary quadratic in u. Before substituting, rewrite every term using the index laws:

4ˣ = (2²)ˣ = (2ˣ)², so 4ˣ = u² when u = 2ˣ.

2ˣ⁺³ = 2³ ×2ˣ = 8u.

3²ˣ⁺¹ = 3 ×(3ˣ)² = 3u² when u = 3ˣ.

Solve for u, then solve aˣ = u for each value. Reject any u ≤0: aˣ is never zero or negative, so such a value gives no x at all.

WORKED EXAMPLE · QUESTION

(a) Solve 4ˣ - 5 ×2ˣ - 6 = 0, giving your answer to 3 significant figures. (b) Solve 3²ˣ⁺¹ - 10 ×3ˣ + 3 = 0.

SOLUTION

(a) Let u = 2ˣ, so 4ˣ = u²: u² - 5u - 6 = 0 ⇒(u - 6)(u + 1) = 0 ⇒u = 6 or u = -1. Now 2ˣ = -1 has no solution, since 2ˣ > 0 for all x. So 2ˣ = 6 ⇒x = log 6/log 2 = 2.5850… x = 2.58

(b) 3²ˣ⁺¹ = 3 ×3²ˣ = 3u² with u = 3ˣ: 3u² - 10u + 3 = 0 ⇒(3u - 1)(u - 3) = 0 ⇒u = ⅓ or u = 3. Both values are positive, so both give a solution: 3ˣ = 3⁻¹ ⇒x = -1, and 3ˣ = 3¹ ⇒x = 1. x = -1 or x = 1

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 = logₐ x, so aʸ = x. Take log_b of both sides: log_b aʸ = log_b x, so ylog_b a = log_b x, giving y = (log_b x)/(log_b a).

Two special cases appear often. Putting x = b gives logₐ b = 1/(log_b a).

Putting a = b² gives log_(b²) x = ½log_b x; for example log₉ x = ½log₃ x.

WORKED EXAMPLE · QUESTION

(a) Find log₅ 12, giving your answer to 3 significant figures. (b) Solve log₃ x = 4logₓ 3, where x > 0 and x ≠1.

SOLUTION

(a) Change to base 10: log₅ 12 = log 12/log 5 = 1.5440… (check: 5^(1.544) ≈12). 1.54

(b) Write logₓ 3 in base 3 and let L = log₃ x: logₓ 3 = (log₃ 3)/(log₃ x) = 1/L, so L = 4/L. Multiply by L (non-zero, since x ≠1): L² = 4, so L = 2 or L = -2. Undo the log: x = 3² = 9 or x = 3⁻² = (1/9). x = 9 or x = (1/9)

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: logₐ P = k ⇒P = aᵏ, and logₐ P = logₐ Q ⇒P = Q. A plain number k can be rewritten as logₐ aᵏ 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. log₃ (x-1)² is defined for every x ≠1, but the original 2log₃(x-1) needs x > 1. The algebra therefore can produce a root that the original equation never allowed.

WORKED EXAMPLE · QUESTION

Solve 2log₃(x - 1) - log₃(x + 5) = 1, where x > 1.

SOLUTION

Power law, then quotient law: log₃ (x-1)² - log₃(x+5) = 1 ⇒log₃ ((x-1)²)/(x+5) = 1.

Remove the log (log₃ P = 1 ⇒P = 3¹): (x-1)² = 3(x+5) ⇒x² - 2x + 1 = 3x + 15 ⇒x² - 5x - 14 = 0.

Solve and check: (x-7)(x+2) = 0, so x = 7 or x = -2. But x = -2 makes x - 1 negative, so log₃(x-1) is undefined. x = 7 only

Check: 2log₃ 6 - log₃ 12 = log₃ (36/12) = log₃ 3 = 1 ✓

WORKED EXAMPLE · QUESTION

(a) Show that log₉ x = ½log₃ x for x > 0. (b) Hence solve log₃ x + log₉ x = 6. (c) Solve 9ˣ - 4 ×3ˣ⁺¹ + 27 = 0. (d) Solve 3ˣ = 2 ×5ˣ⁻¹, giving your answer to 3 significant figures.

SOLUTION

(a) Change of base to base 3: log₉ x = (log₃ x)/(log₃ 9) = (log₃ x)/2 = ½log₃ x ✓

(b) Using (a), let L = log₃ x: L + ½ L = 6 ⇒(3/2) L = 6 ⇒L = 4, so x = 3⁴ = 81. x = 81

(c) 9ˣ = (3ˣ)² and 3ˣ⁺¹ = 3 ×3ˣ; let u = 3ˣ: u² - 12u + 27 = 0 ⇒(u-3)(u-9) = 0 ⇒u = 3 or 9. So 3ˣ = 3 ⇒x = 1, and 3ˣ = 9 ⇒x = 2. x = 1 or x = 2

(d) Take logs, expand, collect: xlog 3 = log 2 + (x-1)log 5 ⇒x(log 3 - log 5) = log 2 - log 5, so

Both brackets are negative, so x is positive; dividing by log 3 - log 5 needs no sign change in an equation. x = 1.79

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 aˣ + 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 rⁿ = 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 eˣ, exponential growth and decay models, and turning y = axⁿ or y = kbˣ 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.

1. Write down log₂ 32 and log₂ (1/32).

2. Show that 3log 2 + log 125 = 3, where log means log₁₀.

3. Solve 3ˣ = 200, giving x to 3 significant figures.

4. Solve 2²ˣ - 3 ×2ˣ + 2 = 0.

5. Solve log₂(x + 3) = 3. Is the answer allowed?

6. Find the least integer n for which 0.8ⁿ < 0.1.

1. 2⁵ = 32, so log₂ 32 = 5; by the reciprocal law log₂ (1/32) = -5.

2. 3log 2 + log 125 = log 8 + log 125 = log 1000 = 3, because 10³ = 1000.

3. x = log 200/log 3 = 4.8227…= 4.82.

4. With u = 2ˣ: u² - 3u + 2 = (u-1)(u-2) = 0, so 2ˣ = 1 or 2, giving x = 0 or x = 1.

5. x + 3 = 2³ = 8, so x = 5. The argument x + 3 = 8 is positive, so yes.

6. Take logs: nlog 0.8 < log 0.1. Since log 0.8 < 0, dividing reverses the inequality: n > log 0.1/log 0.8 = 10.318…, so the least integer is n = 11. (Check: 0.8¹⁰ = 0.107, 0.8¹¹ = 0.0859.)