Year 1 Pure · Chapter 10

Logarithms

Specification: Edexcel Pure 6 — Exponentials and logarithms · AQA section F

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In this chapter

1. What a logarithm is
2. The laws of logarithms
3. Solving exponential equations
4. Solving equations with logs
Practice

Key facts — this chapter

A logarithm answers “what power?”: $\log_a x = y$ means $a^y = x$ (with $a > 0$ , $a \neq 1$ , $x > 0$ ).
So $\log$ is the inverse of raising to a power — the graph of $y = \log_a x$ is $y = a^x$ reflected in $y = x$ .
Laws of logs: $\log_a(xy) = \log_a x + \log_a y$ , $\;\log_a\!\frac{x}{y} = \log_a x - \log_a y$ , $\;\log_a(x^k) = k\log_a x$ .
Useful values: $\log_a a = 1$ and $\log_a 1 = 0$ .
To solve $a^x = b$ , take logs of both sides: $x = \dfrac{\log b}{\log a}$ .
$\log$ usually means base $10$ ; $\ln$ means base $e$ (Chapter 11).

Indices answer “what do I get if I raise $a$ to the power $x$ ?” Logarithms run the question backwards: “what power of $a$ gives me $x$ ?” That single idea — log as the undo button for powers — is the whole chapter.

1. What a logarithm is

Key result

\log_a x = y \quad\Longleftrightarrow\quad a^y = x

Read $\log_a x$ as “the power you raise $a$ to, to get $x$ ”. Because it undoes the exponential, $y = \log_a x$ is the reflection of $y = a^x$ in the line $y = x$ .

The curves y = 2 to the x and y = log base 2 of x, reflections of each other in the line y = x — log is the inverse of the exponential — each graph is the other reflected in y = x.

Worked example 10.1 · Reading logs straight off the definition

Evaluate (a) $\log_2 16$ , (b) $\log_{10} 1000$ , (c) $\log_5 \tfrac{1}{25}$ .

(a) “What power of $2$ gives $16$ ?” $\;2^4 = 16$ , so $\log_2 16 = 4$ .
(b) $10^3 = 1000$ , so $\log_{10} 1000 = 3$ .
(c) $5^{-2} = \tfrac{1}{25}$ , so $\log_5 \tfrac{1}{25} = -2$ .

The graph of y = log x passing through (1, 0), rising slowly, with the y-axis as an asymptote — Every log graph passes through (1, 0) and has the y-axis as an asymptote.

2. The laws of logarithms

The laws of logs are the index laws seen in a mirror — which is exactly why they look so similar.

Key result

\begin{aligned} \log_a(xy) &= \log_a x + \log_a y \\[0.3em] \log_a\!\frac{x}{y} &= \log_a x - \log_a y \\[0.3em] \log_a(x^k) &= k\log_a x \end{aligned}

Plus the two you’ll use constantly: $\log_a a = 1$ and $\log_a 1 = 0$ .

Why it works

Write $x = a^m$ and $y = a^n$ , so $\log_a x = m$ and $\log_a y = n$ . Then $xy = a^m \cdot a^n = a^{m+n}$ by the index laws, and taking the log gives $\log_a(xy) = m + n = \log_a x + \log_a y$ . The product law for logs is just the multiplication law for indices, read backwards.

Pick a base and two numbers. The two lines below are the product law in action: adding the logs gives the same answer as the log of the product.

base

log_b(x × y) = log_b x + log_b y

Worked example 10.2 · Combining logs into one

Write $\log_2 40 - \log_2 5$ as a single logarithm, and hence evaluate it.

The subtraction is the quotient law:
$\log_2 40 - \log_2 5 = \log_2\!\frac{40}{5} = \log_2 8$
And $\log_2 8 = 3$ since $2^3 = 8$ .

Worked example 10.3 · Using the power law

Express $2\log_a 3 + \log_a 2$ as a single logarithm.

The coefficient $2$ goes inside as a power:
$2\log_a 3 = \log_a 3^2 = \log_a 9$
Then add using the product law:
$\log_a 9 + \log_a 2 = \log_a 18$

Common mistake

There is no law for $\log(x + y)$ — it does not equal $\log x + \log y$ (that’s the law for $\log(xy)$ ). Logs only simplify products, quotients and powers, never sums or differences inside the log.

3. Solving exponential equations

When the unknown is stuck in an exponent — like $3^x = 20$ — take the log of both sides to bring it down.

Key result

To solve $a^x = b$ , take logs and use the power law:

\log a^x = \log b \;\;\Longrightarrow\;\; x\log a = \log b \;\;\Longrightarrow\;\; x = \frac{\log b}{\log a}

Any base works (your calculator has $\log$ and $\ln$ ) — just be consistent on both sides.

Worked example 10.4 · An exponential equation

Solve $3^x = 20$ , giving your answer to 3 significant figures.

Take logs of both sides and drop the power down:
$x\log 3 = \log 20 \;\;\Longrightarrow\;\; x = \frac{\log 20}{\log 3}$
Evaluate:
$x = 2.73 \;\,(3\text{ s.f.})$

Worked example 10.5 · The exponent is an expression

Solve $2^{\,x+1} = 7$ , giving your answer to 3 significant figures.

Take logs; the whole exponent $x+1$ comes down:
$(x+1)\log 2 = \log 7 \;\;\Longrightarrow\;\; x + 1 = \frac{\log 7}{\log 2}$
So
$x = \frac{\log 7}{\log 2} - 1 = 1.81 \;\,(3\text{ s.f.})$

4. Solving equations with logs

If the unknown sits inside logs, combine them into one log first, then undo it with the definition.

Worked example 10.6 · A logarithmic equation

Solve $\log_3 x + \log_3(x - 2) = 1$ .

Combine using the product law:
$\log_3\big(x(x-2)\big) = 1$
Undo the log with the definition ( $\log_3 N = 1$ means $N = 3^1$ ):
$x(x-2) = 3 \;\;\Longrightarrow\;\; x^2 - 2x - 3 = 0$
Factorise: $(x-3)(x+1) = 0$ , so $x = 3$ or $x = -1$ .
Check the domain: logs need positive inputs, so $x - 2 > 0$ rules out $x = -1$ (and $x = 3$ works). The solution is $x = 3$ .

Common mistake

Always check your answers against the domain at the end. A log is only defined for a positive argument, so any solution making $x$ or $(x-2)$ zero or negative must be discarded — even though it solved the quadratic.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers. Give non-exact answers to 3 s.f.

Warm-up

1 [3 marks]

Evaluate:

$\log_3 81$
$\log_{10} 0.01$
$\log_7 1$

Show solution

(a) $3^4 = 81$ , so $\log_3 81 = 4$ .

(b) $10^{-2} = 0.01$ , so $\log_{10} 0.01 = -2$ .

(c) $\log_a 1 = 0$ for any base, so $\log_7 1 = 0$ .

2 [3 marks]

Write as a single logarithm: $\log_5 6 + \log_5 4 - \log_5 3$ .

Show solution

Add as a product, subtract as a quotient:

\log_5 6 + \log_5 4 - \log_5 3 = \log_5\!\frac{6 \times 4}{3} = \log_5 8

3 [2 marks]

Express $3\log_a 2$ as a logarithm of a single number.

Show solution

The coefficient becomes a power: $3\log_a 2 = \log_a 2^3 = \log_a 8$ .

4 [2 marks]

Solve $\log_2 x = 5$ .

Show solution

By the definition, $x = 2^5 = 32$ .

Standard

5 [3 marks]

Solve $5^x = 40$ , giving your answer to 3 significant figures.

Show solution

Take logs of both sides:

x\log 5 = \log 40 \;\;\Longrightarrow\;\; x = \frac{\log 40}{\log 5} = 2.29 \;\,(3\text{ s.f.})

6 [4 marks]

Solve $4^{\,2x-1} = 30$ , giving your answer to 3 significant figures.

Show solution

Take logs; the whole exponent comes down:

(2x - 1)\log 4 = \log 30 \;\;\Longrightarrow\;\; 2x - 1 = \frac{\log 30}{\log 4} = 2.4534\ldots

2x = 3.4534\ldots \;\;\Longrightarrow\;\; x = 1.73 \;\,(3\text{ s.f.})

7 [4 marks]

Given that $\log_a x = 2$ and $\log_a y = 5$ , find:

$\log_a (x^2 y)$
$\log_a \dfrac{y}{x}$

Show solution

(a) $\log_a(x^2 y) = 2\log_a x + \log_a y = 2(2) + 5 = 9$ .

(b) $\log_a \dfrac{y}{x} = \log_a y - \log_a x = 5 - 2 = 3$ .

8 [4 marks]

Solve $\log_2(x + 6) = 3 + \log_2 x$ .

Show solution

Write $3 = 3\log_2 2 = \log_2 8$ , then move logs to one side:
$\log_2(x + 6) - \log_2 x = \log_2 8 \;\;\Longrightarrow\;\; \log_2\!\frac{x+6}{x} = \log_2 8$
Equate the arguments:
$\frac{x+6}{x} = 8 \;\;\Longrightarrow\;\; x + 6 = 8x \;\;\Longrightarrow\;\; x = \tfrac{6}{7}$

9 [5 marks]

Solve $\log_4 x + \log_4(x - 6) = 2$ .

Show solution

Combine: $\log_4\big(x(x-6)\big) = 2$ .
Undo the log: $x(x-6) = 4^2 = 16$ , so $x^2 - 6x - 16 = 0$ .
Factorise: $(x - 8)(x + 2) = 0$ , giving $x = 8$ or $x = -2$ .
The argument must be positive, so discard $x = -2$ . Thus $x = 8$ .

Exam style

10 [5 marks]

Solve $3^{2x} - 10\big(3^x\big) + 9 = 0$ .

Show solution

This is a hidden quadratic. Let $u = 3^x$ (so $3^{2x} = u^2$ ):
$u^2 - 10u + 9 = 0$
Factorise: $(u - 1)(u - 9) = 0$ , so $u = 1$ or $u = 9$ .
Convert back to $x$ :
- $3^x = 1 \Rightarrow x = 0$ ;
- $3^x = 9 \Rightarrow x = 2$ .

So $x = 0$ or $x = 2$ .

11 [6 marks]

$\log_2 y = 3 + 2\log_2 x$

Express $y$ in terms of $x$ , with no logarithms. [4]
Hence find $y$ when $x = 4$ . [2]

Show solution

(a)

Turn the right-hand side into a single log: $3 = \log_2 8$ and $2\log_2 x = \log_2 x^2$ , so
$\log_2 y = \log_2 8 + \log_2 x^2 = \log_2(8x^2)$
Equate arguments: $y = 8x^2$ .

(b) When $x = 4$ : $y = 8(4)^2 = 8 \times 16 = 128$ .

12 [5 marks]

The number of bacteria $N$ after $t$ hours is modelled by $N = 200 \times 3^{\,t}$ .

State the number of bacteria at $t = 0$ . [1]
Find the smallest whole number of hours after which the number exceeds $50\,000$ . [4]

Show solution

(a) At $t = 0$ , $N = 200 \times 3^0 = 200$ .

(b) Solve $200 \times 3^t = 50\,000$ :

3^t = 250 \;\;\Longrightarrow\;\; t = \frac{\log 250}{\log 3} = 5.02\ldots

So $N$ passes $50\,000$ during the $6$ th hour. At $t = 5$ it is still below ( $200 \times 3^5 = 48\,600$ ), so the smallest whole number of hours after which it exceeds $50\,000$ is $6$ ( $200 \times 3^6 = 145\,800$ ).