OxfordAQA International AS and A-level Mathematics: Exponentials, logarithms and solving a^x = b (9660)

This guide follows the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section PP1.3 (Exponential and logarithms). The specification places this content in International AS, Unit PSM1 (3.2, PP1), and it applies to International AS exams from May/June 2018 and International A-level exams from May/June 2019.

Use it with the revision notes and the practice questions. The course page is OxfordAQA A-level Mathematics, and the printable checklist lets you tick off each outcome. If you are not sure where to start, take a free 10-minute diagnostic first.

What PP1.3 asks you to do

Specification item	What you must be able to do
y = a^x and its graph	Sketch the graph for a > 1 and 0 < a < 1, using the laws of indices where they help
Logarithms and the laws of logarithms	Use logₐx + logₐy = logₐ(xy), logₐx − logₐy = logₐ(x/y) and k logₐx = logₐ(x^k)
Equivalence of y = a^x and x = logₐy	Move between index form and logarithm form
Solving a^x = b	Solve equations such as 3^(2x) = 2, including by using the logarithm function on your calculator

1. The graph of y = a^x

Take a > 0 and a ≠ 1. The graph of y = a^x always has these features.

• The point (0, 1) is always on the curve, since a⁰ = 1.
• It lies above the x-axis for every x, because a^x is never zero or negative.
• The x-axis is an asymptote: the curve gets closer to y = 0 but never reaches it.
• If a > 1 the curve rises, and it rises faster as x increases. If 0 < a < 1 it falls.

Here are values for y = 4^x and y = (1/4)^x.

x	−2	−1	0	1	2
4^x	1/16	1/4	1	4	16
(1/4)^x	16	4	1	1/4	1/16

The second row is the first row reversed. That is because (1/4)^x = 4^(−x), so the graph of y = (1/4)^x is the reflection of y = 4^x in the y-axis.

Sketch description for y = 4^x: start just above the negative x-axis on the left, cross the y-axis at (0, 1), then climb steeply, passing (1, 4) and (2, 16). Label the intercept (0, 1) and say “asymptote y = 0”.

Laws of indices you will use

a^m × a^n = a^(m+n)        a^m ÷ a^n = a^(m−n)
(a^m)^n = a^(mn)           a^(−n) = 1/a^n
a^0 = 1                    a^(1/n) is the n-th root of a

Example 1. Solve 125^(x+1) = 25^(2x−1).

Write both sides as powers of 5.

125 = 5³ and 25 = 5²
5^(3(x+1)) = 5^(2(2x−1))
3(x+1) = 2(2x−1)
3x + 3 = 4x − 2
x = 5

Check: 125⁶ = 5¹⁸ and 25⁹ = 5¹⁸. This only works when both sides can be written with the same base. When they cannot, you need logarithms (section 4).

2. Logarithms and the equivalence with indices

The logarithm answers the question “what power?”. The two statements below say the same thing.

y = a^x   ⇔   x = logₐ y        (a > 0, a ≠ 1, y > 0)

So log₂32 = 5 because 2⁵ = 32. The logarithm is the inverse of the exponential, so the graph of y = logₐx is the graph of y = a^x reflected in the line y = x. You can only take the logarithm of a positive number.

Two results follow straight from a¹ = a and a⁰ = 1:

logₐ a = 1        logₐ 1 = 0

Example 2. Evaluate (a) log₅(1/25), (b) log₉27, (c) log₈4.

(a) Let log₅(1/25) = k. Then 5^k = 1/25 = 5⁻², so k = −2.

(b) Let log₉27 = k. Then 9^k = 27, so 3^(2k) = 3³. Hence 2k = 3 and k = 3/2.

(c) Let log₈4 = k. Then 8^k = 4, so 2^(3k) = 2². Hence 3k = 2 and k = 2/3.

The method is always the same: write the logarithm equal to k, convert to index form, then match powers of a common base.

Example 3. Find x when (a) log_x 81 = 4, (b) log₂x = −3.

(a) x⁴ = 81, so x = 3 (a base must be positive).

(b) x = 2⁻³ = 1/8.

3. The laws of logarithms

All three laws need the same base on every term.

logₐ x + logₐ y = logₐ(xy)
logₐ x − logₐ y = logₐ(x/y)
k logₐ x = logₐ(x^k)

Why they hold: if x = a^p and y = a^q, then xy = a^(p+q), so logₐ(xy) = p + q = logₐx + logₐy. The other laws come from the same idea with a^p ÷ a^q and (a^p)^k.

Example 4. Evaluate without a calculator.

(a) log₆4 + log₆9 = log₆36 = 2

(b) log₃54 − log₃2 = log₃27 = 3

(c) 2log₅10 − log₅4 = log₅100 − log₅4 = log₅25 = 2

Example 5. Suppose logₐ2 = p and logₐ3 = q. Write logₐ12 and logₐ(9/8) in terms of p and q.

logₐ12 = logₐ(4 × 3) = 2logₐ2 + logₐ3 = 2p + q
logₐ(9/8) = 2logₐ3 − 3logₐ2 = 2q − 3p

Split the number into factors that are powers of 2 and 3 first, then apply the laws.

Example 6. Solve log₂x + log₂(x − 2) = 3.

log₂[x(x − 2)] = 3
x(x − 2) = 2³ = 8
x² − 2x − 8 = 0
(x − 4)(x + 2) = 0

The roots are x = 4 and x = −2. Because log₂x needs x > 0, discard x = −2. The solution is x = 4. Check: log₂4 + log₂2 = 2 + 1 = 3.

4. Solving a^x = b

When the bases will not match, take logarithms of both sides. You may choose any base for the logarithms provided both sides use the same one; the calculator’s logarithm function then gives numerical values. The key step is the third law: it brings the unknown power down in front.

a^x = b
x log a = log b
x = log b / log a

Example 7 (the specification’s own type). Solve 3^(2x) = 2.

2x log 3 = log 2
x = log 2 / (2 log 3) = 0.6931 / 2.1972
x = 0.315   (3 s.f.)

Check: 3^(0.6309) = 2.

Example 8. Solve 7^(2x−1) = 30.

(2x − 1) log 7 = log 30
2x − 1 = log 30 / log 7 = 1.7479
2x = 2.7479
x = 1.37   (3 s.f.)

Divide first, then rearrange. Avoid rounding 1.7479 early; carry the unrounded calculator value to the final line.

Example 9. Solve 2^(x+1) = 3^x.

(x + 1) log 2 = x log 3
x log 2 + log 2 = x log 3
log 2 = x (log 3 − log 2)
x = log 2 / (log 3 − log 2) = 1.71   (3 s.f.)

When the unknown is on both sides, collect the x terms on one side and factor.

Example 10. Solve 0.8^x = 0.3.

x log 0.8 = log 0.3
x = log 0.3 / log 0.8 = 5.40   (3 s.f.)

Both logarithms are negative, so the quotient is positive. That agrees with the graph: 0.8^x falls, and it equals 0.3 at a positive x.

Using this in context

Example 11. A colony starts at 800 cells and grows by 15% each hour, so after t hours it has 800 × 1.15^t cells. When does it reach 5000?

800 × 1.15^t = 5000
1.15^t = 6.25
t = log 6.25 / log 1.15 = 13.1 hours

Example 12. A value falls by 10% each year from a start of 1. Find the first whole number of years n for which 0.9^n < 0.2.

n log 0.9 < log 0.2

Now divide by log 0.9. It is negative, so the inequality reverses.

n > log 0.2 / log 0.9 = 15.28

The smallest whole number is n = 16. Check: 0.9¹⁵ = 0.206 (not below 0.2) and 0.9¹⁶ = 0.185 (below 0.2).

Common errors

• Splitting log(x + y) into log x + log y. Only a product inside the bracket splits, as in logₐ(xy).
• Writing log 6 − log 2 = log 4. Subtraction of logarithms gives log(6/2) = log 3.
• Writing (log x)² = 2 log x. The third law applies to log(x²), not to (log x)².
• Combining logarithms with different bases without changing them first.
• Forgetting to check log equations. A root that makes an argument zero or negative must be rejected.
• Rounding part-way through a^x = b, which can change the third significant figure.
• Dividing an inequality by a negative logarithm and keeping the sign. If the base is between 0 and 1, the logarithm is negative and the sign reverses.
• Saying a^x can be negative or zero. It cannot, so an equation such as 3^x = −2 has no solution.

Where to go next

Try the practice questions and then recheck with the revision notes. For the pure maths in the neighbouring Unit P1, see the Unit P1 study guide.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2: section PP1.3, Exponential and logarithms. Published by OxfordAQA for International AS exams from May/June 2018 and International A-level exams from May/June 2019.