Pure mathematics

Exponentials and Logarithms

Year 12 · Year 13

✓By the end of this lesson students will be able to understand and use the functions e^x and ln x.
✓By the end of this lesson students will be able to apply the laws of logarithms to simplify expressions and solve equations.
✓By the end of this lesson students will be able to solve problems involving exponential growth and decay models.
✓By the end of this lesson students will be able to transform data to linear form using logarithms and interpret log-linear graphs.

Key concepts

The Exponential Function e^x

The exponential function, denoted as e^x, is a fundamental function in mathematics. 'e' is Euler's number, an irrational constant approximately equal to 2.71828. The function y = e^x has a domain of all real numbers and a range of y > 0. Its graph passes through (0,1) and increases rapidly. It is unique in that its gradient at any point is equal to its y-value at that point.

The Natural Logarithm ln x

The natural logarithm, denoted as ln x (or log_e x), is the inverse function of e^x. This means that if y = ln x, then x = e^y. The domain of ln x is x > 0, and its range is all real numbers. Its graph passes through (1,0) and increases. Key inverse properties are e^(ln x) = x for x > 0, and ln(e^x) = x for all real x.

y = ln x ⇔ x = e^y

Laws of Logarithms

Logarithms follow specific rules that allow for simplification and manipulation of expressions. These laws are crucial for solving exponential and logarithmic equations. The base 'a' must be a positive number not equal to 1. For natural logarithms, 'a' is 'e'.

log_a (xy) = log_a x + log_a y log_a (x/y) = log_a x - log_a y log_a (x^n) = n log_a x log_a 1 = 0 log_a a = 1 log_a x = (log_b x) / (log_b a)

Exponential Models

Exponential functions are used to model real-world phenomena involving rapid growth or decay, such as population growth, radioactive decay, or compound interest. The general form for continuous growth/decay is P = P_0 e^(kt), where P is the quantity at time t, P_0 is the initial quantity, and k is the growth/decay constant (k > 0 for growth, k < 0 for decay).

P = P_0 e^(kt)

Log-linear Graphs

Some non-linear relationships can be transformed into linear relationships by taking logarithms of one or both variables. This allows us to use linear regression techniques to find the constants in the original non-linear model. Two common forms are y = ax^n and y = ab^x.

For y = ax^n: log y = n log x + log a (linear form Y = mX + c where Y=log y, X=log x, m=n, c=log a) For y = ab^x: log y = x log b + log a (linear form Y = mX + c where Y=log y, X=x, m=log b, c=log a)

Key facts to remember

1Euler's number, e, is an irrational constant approximately 2.71828.
2The natural logarithm, ln x, is equivalent to log_e x.
3e^(ln x) = x for x > 0, and ln(e^x) = x for all real x.
4Laws of logarithms: log(AB) = log A + log B, log(A/B) = log A - log B, log(A^n) = n log A.
5log_a 1 = 0 and log_a a = 1 for any valid base a.
6The domain of log_a x is x > 0.
7The change of base formula is log_a x = (log_b x) / (log_b a), commonly used with b=e (ln) or b=10 (log_10).
8Exponential growth/decay models are typically of the form P = P_0 e^(kt) or P = P_0 r^t.

Worked examples

Example 1

Solve the equation 5^(2x-1) = 12, giving your answer to 3 significant figures.

ITake the natural logarithm (ln) of both sides: ln(5^(2x-1)) = ln(12)

IIApply the power law of logarithms: (2x-1)ln(5) = ln(12)

IIIExpand the left side: 2x ln(5) - ln(5) = ln(12)

IVRearrange to isolate the term with x: 2x ln(5) = ln(12) + ln(5)

VSolve for x: x = (ln(12) + ln(5)) / (2 ln(5))

VICalculate the value: x = (2.4849 + 1.6094) / (2 * 1.6094) = 4.0943 / 3.2188 = 1.2719...

VIIRound to 3 significant figures.

Answer

x = 1.27

You could use log base 10 instead of ln, the result will be the same.

Example 2

The number of bacteria, N, in a culture at time t hours is modelled by the equation N = N_0 e^(kt). Initially, there are 200 bacteria. After 4 hours, the number of bacteria has increased to 800. a) Find the values of N_0 and k. b) Find the time it takes for the number of bacteria to reach 5000.

Ia) Find N_0 and k:

II At t=0, N=200. Substitute into N = N_0 e^(kt): 200 = N_0 e^(k*0) ⇒ 200 = N_0 * 1 ⇒ N_0 = 200.

III So the model is N = 200e^(kt).

IV At t=4, N=800. Substitute into the model: 800 = 200e^(4k).

V Divide by 200: 4 = e^(4k).

VI Take the natural logarithm of both sides: ln(4) = ln(e^(4k)).

VII Simplify: ln(4) = 4k.

VIII Solve for k: k = ln(4) / 4 = 1.38629... / 4 = 0.34657...

9 Round k to 3 significant figures: k = 0.347.

10b) Find the time for N=5000:

11 Use the model N = 200e^(0.34657t) (using the unrounded k for accuracy).

12 Set N=5000: 5000 = 200e^(0.34657t).

13 Divide by 200: 25 = e^(0.34657t).

14 Take the natural logarithm of both sides: ln(25) = ln(e^(0.34657t)).

15 Simplify: ln(25) = 0.34657t.

16 Solve for t: t = ln(25) / 0.34657 = 3.21887... / 0.34657 = 9.287... hours.

17 Round t to 3 significant figures.

Answer

a) N_0 = 200, k = 0.347 (3 s.f.) b) t = 9.29 hours (3 s.f.)

Always use the unrounded value of 'k' in subsequent calculations to maintain accuracy, only rounding the final answer.

Example 3

Experimental data suggests a relationship between variables x and y of the form y = ax^b. A graph of log_10 y against log_10 x is plotted and found to be a straight line passing through the points (1, 2.5) and (3, 5.5). Find the values of a and b, giving your answers to 3 significant figures.

IThe relationship y = ax^b can be transformed by taking log_10 of both sides:

II log_10 y = log_10 (ax^b)

III log_10 y = log_10 a + log_10 (x^b)

IV log_10 y = b log_10 x + log_10 a.

VThis is in the form Y = mX + c, where Y = log_10 y, X = log_10 x, m = b, and c = log_10 a.

VICalculate the gradient (m = b) using the given points (X1, Y1) = (1, 2.5) and (X2, Y2) = (3, 5.5):

VII b = (Y2 - Y1) / (X2 - X1) = (5.5 - 2.5) / (3 - 1) = 3 / 2 = 1.5.

VIIINow find the y-intercept (c = log_10 a) using the equation Y = bX + c and one of the points, e.g., (1, 2.5):

9 2.5 = 1.5 * 1 + c

10 2.5 = 1.5 + c

11 c = 1.

12Since c = log_10 a, we have log_10 a = 1.

13To find a, raise 10 to the power of both sides: a = 10^1 = 10.

14State the values of a and b, rounded to 3 significant figures if necessary (here they are exact).

Answer

a = 10.0 (3 s.f.), b = 1.50 (3 s.f.)

Ensure you correctly identify which part of the linear equation corresponds to 'a' and 'b' from the original model.

Common mistakes

✗Incorrectly applying logarithm laws, e.g., assuming log(A+B) = log A + log B or log(A-B) = log A - log B.
✗Forgetting that the argument of a logarithm must be positive (i.e., ln x is only defined for x > 0).
✗Confusing log_10 (common logarithm) with ln (natural logarithm) on a calculator, leading to incorrect values.
✗Errors with signs in exponential models, particularly using a positive 'k' for decay or a negative 'k' for growth.
✗Misinterpreting the gradient or y-intercept in log-linear graphs, e.g., stating the intercept is 'a' instead of 'log a'.

Exam tips

★Always show full step-by-step working when applying logarithm laws to simplify expressions or solve equations.
★Be careful with calculator use; ensure you are using the correct 'e^x' and 'ln' buttons, and round only at the final step to the required accuracy.
★When solving equations involving logarithms, always check your solutions against the domain of the original logarithmic expressions.
★For log-linear graphs, clearly state the linear form (Y = mX + c) you are using and identify what Y, X, m, and c represent in terms of the original variables and constants.

Ready to practise?

Try a problem on this topic

Snap a photo or type a question — get step-by-step working instantly.

Solve a problem →← Maths curriculum