Simplify _3 81 - _3 9. [2 marks]

Express 2_a 3 + _a 5 as a single logarithm. [2 marks]

SQA SQA Higher 2018-style practice: Experimental data for two quantities x and y is believed to follow a law of the form y = k x^n. When _10 y is plotted against _10 x, a straight line of gradient 1.5 passing through (0, 0.6) is obtained. Determine the values of k and n, and hence express y in terms of x.

Take logs of y = k x^n: _10 y = _10 k + n _10 x, a straight line with gradient n and intercept _10 k (1 mark). The gradient is n, so n = 1.5 (1 mark). The intercept on the _10 y axis is _10 k = 0.6, so k = 10^0.6 ≈ 3.98 (2 marks). Therefore y = 3.98 x^1.5 (1 mark). Markers reward taking logs to linear form, identifying n from the gradient, recovering k by raising 10 to the intercept, and the final law.

ScotlandMathsSyllabus dot point

How do the laws of logarithms let you solve equations with the unknown in the power, and how do logs reveal hidden relationships in data?

The laws of logarithms and their use in simplifying expressions and solving equations, the relationship between exponential and logarithmic form, and the use of logarithms to find the parameters in experimental laws of the form y equals k x to the n and y equals a b to the x.

A focused answer to the SQA Higher Mathematics exponentials and logarithms content, covering the laws of logarithms, the link between exponential and logarithmic form, solving equations with the unknown in the power, and using logs to find the parameters in experimental power and exponential laws.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Logs and exponentials
The laws of logarithms
Straightening experimental data
Examples in context
Try this

What this dot point is asking

The SQA wants you to use the laws of logarithms to simplify expressions and solve equations, move between exponential and logarithmic form, and use logarithms to straighten experimental data so you can find the constants in laws of the form $y = kx^n$ and $y = ab^x$ .

Logs and exponentials

Because they are inverse operations, an exponential and a logarithm undo each other: $a^{\log_a x} = x$ and $\log_a(a^x) = x$ . The natural logarithm $\ln x = \log_e x$ uses the base $e \approx 2.718$ , and obeys the same laws.

Convert between forms

Three conversion tasks illustrate the equivalence between exponential and logarithmic statements.

Step 1: Convert from exponential to logarithmic form

The statement $2^5 = 32$ says "the base $2$ raised to the power $5$ gives $32$ ". The logarithm captures that power: $\log_2 32 = 5$ . In general, $a^x = b$ becomes $\log_a b = x$ .

Step 2: Convert from logarithmic to exponential form

The statement $\log_3 81 = 4$ says "the base $3$ must be raised to the power $4$ to give $81$ ". Writing it back as an exponential: $3^4 = 81$ . Conversion in this direction is the key move for solving logarithmic equations.

Step 3: Evaluate a logarithm from scratch

$\log_5 125$ asks the question "to what power must $5$ be raised to produce $125$ ?" Since $5^1 = 5$ , $5^2 = 25$ and $5^3 = 125$ , the answer is $3$ , so $\log_5 125 = 3$ .

Final answer: $\log_2 32 = 5$ ; $\log_3 81 = 4$ means $3^4 = 81$ ; $\log_5 125 = 3$ .

The laws of logarithms

Straightening experimental data

The trick is that taking logs converts a power law or an exponential law into a straight line, which experimenters can fit easily. For $y = kx^n$ , taking logs of both sides gives $\log y = \log k + n \log x$ , so a plot of $\log y$ against $\log x$ is linear with gradient $n$ and intercept $\log k$ . For $y = ab^x$ , taking logs gives $\log y = \log a + x \log b$ , so $\log y$ against $x$ is linear with gradient $\log b$ and intercept $\log a$ .

Find the constants for

y = kx^n

from a log-log line of gradient

2

and intercept

0.7

A plot of $\log_{10} y$ against $\log_{10} x$ gives a straight line with gradient $2$ and intercept $0.7$ . Find $k$ and $n$ , and write the law.

Step 1: Explain why the log-log plot is linear

Taking $\log_{10}$ of both sides of $y = kx^n$ gives:

\log_{10} y = \log_{10} k + n \log_{10} x

This has the form $Y = c + mX$ where $Y = \log_{10} y$ and $X = \log_{10} x$ . So a log-log plot is linear with gradient $n$ and vertical intercept $\log_{10} k$ . The power law has been "straightened" into a line.

Step 2: Read off the constants from the line

The gradient of the line equals $n$ , so $n = 2$ . The intercept equals $\log_{10} k$ , so $\log_{10} k = 0.7$ . Raise the base $10$ to this power to recover $k$ :

k = 10^{0.7} \approx 5

Step 3: State the law

y \approx 5x^2

Final answer: $n = 2$ and $k \approx 5$ , giving the law $y \approx 5x^2$ . Always undo the log on the intercept by raising $10$ to that power to find the multiplier $k$ .

Solve a logarithmic equation by combining terms

Solve $\log_2 x + \log_2(x - 6) = 4$ .

Step 1: Combine the logarithms into a single term

The addition law $\log_a x + \log_a y = \log_a(xy)$ lets us merge the left-hand side. Both logs have the same base $2$ , so:

\log_2\!\big(x(x - 6)\big) = 4

Combining first is essential: you cannot convert to exponential form while there are two separate logarithms.

Step 2: Convert to exponential form and solve

Rewrite $\log_2(\ldots) = 4$ as an exponential statement:

x(x - 6) = 2^4 = 16 \implies x^2 - 6x - 16 = 0

Factorise: $(x - 8)(x + 2) = 0$ , so $x = 8$ or $x = -2$ .

Step 3: Check the domain and reject invalid solutions

Both original logarithms require their arguments to be positive. For $\log_2 x$ we need $x > 0$ , and for $\log_2(x - 6)$ we need $x > 6$ . The solution $x = -2$ fails both conditions and must be rejected.

Final answer: The solution is $x = 8$ .

Examples in context

Radioactive decay and population growth both follow $y = ab^x$ laws. Suppose a sample's activity is recorded and a plot of $\log_{10} y$ against time $x$ gives a straight line with gradient $-0.1$ and intercept $2$ . Then $\log_{10} a = 2$ so $a = 100$ , and $\log_{10} b = -0.1$ so $b = 10^{-0.1} \approx 0.79$ . The law $y = 100(0.79)^x$ says the initial activity is $100$ units and about $21\%$ is lost each time unit, exactly the kind of model a physicist extracts from raw readings.

Try this

Q1. Simplify $\log_3 81 - \log_3 9$ . [2 marks]

Cue. $\log_3\!\left(\dfrac{81}{9}\right) = \log_3 9 = 2$ .

Q2. Solve $2^x = 50$ , giving your answer to two decimal places. [3 marks]

Cue. $x = \dfrac{\log 50}{\log 2} \approx 5.64$ .

Q3. Express $2\log_a 3 + \log_a 5$ as a single logarithm. [2 marks]

Cue. $\log_a 3^2 + \log_a 5 = \log_a 9 + \log_a 5 = \log_a 45$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20214 marksSolve the equation

\log_4(x + 1) + \log_4(x - 2) = 1

, where

x > 2

Show worked answer →

Combine the logs using the addition law: $\log_4\big((x + 1)(x - 2)\big) = 1$ (1 mark).

Rewrite in exponential form: $(x + 1)(x - 2) = 4^1 = 4$ (1 mark).

Expand and solve: $x^2 - x - 2 = 4$ , so $x^2 - x - 6 = 0$ , giving $(x - 3)(x + 2) = 0$ and $x = 3$ or $x = -2$ (1 mark).

Reject $x = -2$ because the logs require $x > 2$ (and the arguments must be positive), so $x = 3$ (1 mark). Markers reward combining the logs, converting to exponential form, solving the quadratic, and discarding the invalid root.

SQA Higher 20185 marksExperimental data for two quantities

x

and

y

is believed to follow a law of the form

y = k x^n

. When

\log_{10} y

is plotted against

\log_{10} x

, a straight line of gradient

1.5

passing through

(0, 0.6)

is obtained. Determine the values of

k

and

n

, and hence express

y

in terms of

x

Show worked answer →

Take logs of $y = k x^n$ : $\log_{10} y = \log_{10} k + n \log_{10} x$ , a straight line with gradient $n$ and intercept $\log_{10} k$ (1 mark).

The gradient is $n$ , so $n = 1.5$ (1 mark).

The intercept on the $\log_{10} y$ axis is $\log_{10} k = 0.6$ , so $k = 10^{0.6} \approx 3.98$ (2 marks).

Therefore $y = 3.98 x^{1.5}$ (1 mark). Markers reward taking logs to linear form, identifying $n$ from the gradient, recovering $k$ by raising $10$ to the intercept, and the final law.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)