Logarithms and the Laws of Logarithms

A2 · P3 · 10 min

A logarithm answers the question "what power?". $\log_2 8 = 3$ because $2^3 = 8$ . Every law of logarithms is a law of indices in disguise, and once you see that, the laws stop being things to memorise and become things you can rebuild in seconds. Logarithms are used throughout P3: to solve equations with the unknown in a power, to turn curved data into straight lines, and in every integral that produces $\ln$ .

Logarithms are powers

Definition

For a base $b > 0$ with $b \ne 1$ , and $x > 0$ ,

\log_b x = y \iff b^y = x.

$\log_b x$ is the power to which $b$ must be raised to give $x$ .

Read it aloud as "log base $b$ of $x$ is the power you put on $b$ to get $x$ ". Some values to make it concrete:

Index form	Log form
$2^5 = 32$	$\log_2 32 = 5$
$10^{-2} = 0.01$	$\log_{10} 0.01 = -2$
$9^{1/2} = 3$	$\log_9 3 = \tfrac{1}{2}$
$5^0 = 1$	$\log_5 1 = 0$
$e^1 = e$	$\ln e = 1$

Two bases have their own notation:

$\lg x$ or $\log x$ means $\log_{10} x$ (the common logarithm).
$\ln x$ means $\log_e x$ (the natural logarithm), where $e \approx 2.71828$ ; see the exponential function and natural logarithm.

Since $b^y$ is always positive, you can only take the logarithm of a positive number. $\log_2 0$ and $\log_2(-4)$ do not exist. This restriction matters every time you solve a log equation.

Key result

Immediate consequences of the definition, for $b > 0$ , $b \ne 1$ :

\log_b 1 = 0, \qquad \log_b b = 1, \qquad \log_b(b^k) = k, \qquad b^{\log_b x} = x \ \ (x > 0).

The last two say that " $\log_b$ " and " $b$ to the power" undo each other: they are inverse functions.

Evaluating logarithms

Find the exact values of (a) $\log_3 \tfrac{1}{81}$ , (b) $\log_8 2$ , (c) $\log_4 8$ .

Solution

Write each number as a power of the base.

(a) $\tfrac{1}{81} = 3^{-4}$ , so $\log_3 \tfrac{1}{81} = -4$ .

(b) $2 = 8^{1/3}$ , since $\sqrt[3]{8} = 2$ . So $\log_8 2 = \tfrac{1}{3}$ .

The three laws

Key result

For $p > 0$ , $q > 0$ , any real $k$ , and a fixed base $b$ :

\log_b(pq) = \log_b p + \log_b q

\log_b\left(\frac{p}{q}\right) = \log_b p - \log_b q

\log_b(p^k) = k\log_b p

Special cases: $\log_b\left(\dfrac{1}{q}\right) = -\log_b q$ and $\log_b \sqrt{p} = \tfrac{1}{2}\log_b p$ .

The laws come from the laws of indices

Let $\log_b p = m$ and $\log_b q = n$ , so $p = b^m$ and $q = b^n$ .

Multiplication: $pq = b^m b^n = b^{m + n}$ , so $\log_b(pq) = m + n = \log_b p + \log_b q$ .

Division: $\dfrac{p}{q} = b^{m - n}$ , so $\log_b\dfrac{p}{q} = m - n$ .

Powers: $p^k = (b^m)^k = b^{km}$ , so $\log_b(p^k) = km = k\log_b p$ .

So "multiplying numbers adds their logs" because "multiplying powers adds the indices". This is exactly why logs were invented: they turn multiplication into addition.

Watch out

The laws do not say anything about the log of a sum. $\log(p + q)$ is not $\log p + \log q$ , and it cannot be simplified. Similarly $(\log p)^2 \ne 2\log p$ and $\dfrac{\log p}{\log q} \ne \log\dfrac{p}{q}$ .

Tip

The syllabus excludes the change-of-base formula $\log_b x = \dfrac{\ln x}{\ln b}$ , so you will not be asked to use or prove it. It is still a handy way to check an answer on a calculator.

Simplifying and combining

To write several logs as one, first move every coefficient inside as a power (the power law), then combine with the product and quotient laws.

Combining into a single logarithm

Express $2\ln 3 - \ln 6 + \tfrac{1}{2}\ln 16$ as a single logarithm in its simplest form.

Solution

Powers first: $2\ln 3 = \ln 9$ and $\tfrac{1}{2}\ln 16 = \ln 16^{1/2} = \ln 4$ .

\ln 9 - \ln 6 + \ln 4 = \ln\frac{9 \times 4}{6} = \ln 6.

Expressing in terms of given logs

Given that $\ln x = p$ and $\ln y = q$ , express $\ln\left(\dfrac{x^3\sqrt{y}}{e^2}\right)$ in terms of $p$ and $q$ .

Solution

\ln\left(\frac{x^3\sqrt{y}}{e^2}\right) = \ln(x^3) + \ln(y^{1/2}) - \ln(e^2) = 3\ln x + \tfrac{1}{2}\ln y - 2 = 3p + \tfrac{1}{2}q - 2.

(Using $\ln(e^2) = 2$ .)

Solving equations involving logarithms

There are two standard moves:

Combine the logs on each side into a single log, then
Undo the log by raising the base to each side: $\log_b(\ldots) = c \Rightarrow (\ldots) = b^c$ , or, if both sides are single logs to the same base, $\log_b A = \log_b B \Rightarrow A = B$ .

Then solve the resulting ordinary equation, and check every solution in the original equation, because each log needs a positive argument.

Solving a logarithmic equation

Note the domain: every expression inside a log must be positive.
Use the laws to combine the logs into one log on each side (move numbers like $1 = \log_b b$ or $2 = \log_b b^2$ into log form if needed).
Remove the logs: $\log_b A = c \Rightarrow A = b^c$ , or $\log_b A = \log_b B \Rightarrow A = B$ .
Solve the resulting algebraic equation.
Reject any solution that makes an argument of a log zero or negative.

Two logs equal to a number

Solve $\log_2(x + 3) + \log_2 x = 2$ .

Solution

Domain: $x + 3 > 0$ and $x > 0$ , so $x > 0$ .

\log_2\big(x(x + 3)\big) = 2 \quad\Rightarrow\quad x^2 + 3x = 2^2 = 4 \quad\Rightarrow\quad x^2 + 3x - 4 = 0 \quad\Rightarrow\quad (x + 4)(x - 1) = 0.

$x = -4$ is outside the domain ( $\log_2(-4)$ is undefined), so reject it. The solution is $x = 1$ .

Check: $\log_2 4 + \log_2 1 = 2 + 0 = 2$ .

Logs on both sides

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

Solution

Domain: $x > 2$ .

\log_3\frac{x^2}{x - 2} = 2 \quad\Rightarrow\quad \frac{x^2}{x - 2} = 9 \quad\Rightarrow\quad x^2 = 9x - 18 \quad\Rightarrow\quad x^2 - 9x + 18 = 0.

$(x - 3)(x - 6) = 0$ , so $x = 3$ or $x = 6$ . Both are greater than $2$ , so both are valid.

Check $x = 3$ : $2\log_3 3 - \log_3 1 = 2 - 0 = 2$ . Check $x = 6$ : $\log_3 36 - \log_3 4 = \log_3 9 = 2$ .

An equation with no solution

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

Solution

Domain: $x > \tfrac{1}{2}$ .

Write $1 = \ln e$ : $\ln(2x - 1) = \ln e + \ln x = \ln(ex)$ . So

2x - 1 = ex \quad\Rightarrow\quad x(2 - e) = 1 \quad\Rightarrow\quad x = \frac{1}{2 - e} \approx -1.39.

This is not greater than $\tfrac{1}{2}$ (it is negative, so $\ln x$ does not even exist). The equation has no solution.

Watch out

"Undoing" the log must be done to the whole side. From $\ln(2x - 1) = 1 + \ln x$ you cannot write $2x - 1 = e + x$ . First combine the right-hand side into a single log, $\ln(ex)$ , and only then remove the logs.

Rearranging relationships

Questions often give a relationship between $x$ and $y$ in log form and ask for $y$ in terms of $x$ with no logs. Combine into a single log on each side, then remove the logs.

Removing logs from a relationship

Given that $\ln(y + 1) - \ln y = 2\ln x + \ln 3$ , express $y$ in terms of $x$ .

Solution

\ln\frac{y + 1}{y} = \ln(3x^2) \quad\Rightarrow\quad \frac{y + 1}{y} = 3x^2 \quad\Rightarrow\quad y + 1 = 3x^2y.

Collect the $y$ terms: $1 = y(3x^2 - 1)$ , so

y = \frac{1}{3x^2 - 1}.

Common mistakes

Watch out

Splitting a log of a sum. $\ln(x + 3) \ne \ln x + \ln 3$ . Only products, quotients and powers can be split.

Watch out

Applying the power law to only part of a term. $\log(2x)^3 = 3\log(2x)$ , but $\log(2x^3) = \log 2 + 3\log x$ . Brackets decide what the power applies to.

Watch out

Keeping solutions outside the domain. Every argument of every log in the original equation must be positive. Check each solution and reject with a reason, for example " $x = -4$ rejected since $\log_2(-4)$ is undefined".

Watch out

Treating $\ln$ as a number that can be cancelled. $\dfrac{\ln 8}{\ln 2}$ is $3$ (because $\ln 8 = 3\ln 2$ ), not $\ln 4$ .

Exam tip

"Exact value" means leave answers like $\ln 6$ , $\tfrac{1}{2 - e}$ or $\dfrac{\ln 12}{\ln(16/3)}$ ; do not convert to decimals.
Show the combination step explicitly, such as $\log_2\big(x(x + 3)\big) = 2$ , then the line without logs. Examiners award a method mark for each.
When rejecting a root, state the reason. A bare " $x = 1$ " after finding $x = -4$ and $x = 1$ often loses the final mark.
$\ln$ and $\lg$ are both on your calculator; check numerical answers by substituting back.

Summary

$\log_b x = y \iff b^y = x$ ; logs exist only for positive $x$ .
$\log_b 1 = 0$ , $\log_b b = 1$ , $\log_b b^k = k$ , $b^{\log_b x} = x$ .
$\log(pq) = \log p + \log q$ , $\log\dfrac{p}{q} = \log p - \log q$ , $\log p^k = k\log p$ ; these come from the laws of indices.
There is no law for $\log(p + q)$ .
To solve: combine to a single log on each side, remove the logs, solve, reject anything outside the domain.
$\ln$ is $\log_e$ , $\lg$ is $\log_{10}$ ; change of base is not examined.

Practice

Question

Find the exact values of $\log_5 125$ , $\log_2 \tfrac{1}{16}$ , $\log_9 27$ and $\log_{1/2} 8$ .
Show that $3\lg 2 + \lg 5 - \lg 4 = 1$ .
Given $\ln x = p$ and $\ln y = q$ , express $\ln\left(\dfrac{ex^2}{\sqrt{y}}\right)$ in terms of $p$ and $q$ .
Simplify $\log_3 54 - \log_3 2$ .
Solve $\log_2(x + 1) - \log_2(x - 2) = 2$ .
Solve $\lg x + \lg(x - 3) = 1$ .
Solve $\ln(x + 2) = 2\ln x + \ln 2$ , giving your answer in exact form.
Express $y$ in terms of $x$ , without logarithms: (a) $\ln y = 2 + 3\ln x$ ; (b) $\log_2 y = 3 - \log_2 x$ .
Given that $2\ln(x - y) = \ln x + \ln y$ , where $x > y > 0$ , show that $x^2 - 3xy + y^2 = 0$ , and hence find the exact value of $\dfrac{x}{y}$ .

Answers

$125 = 5^3$ , so $3$ . $\tfrac{1}{16} = 2^{-4}$ , so $-4$ . $27 = 9^{3/2}$ , so $\tfrac{3}{2}$ . $8 = \left(\tfrac{1}{2}\right)^{-3}$ , so $-3$ .
$\lg 8 + \lg 5 - \lg 4 = \lg\dfrac{8 \times 5}{4} = \lg 10 = 1$ .
$\ln e + 2\ln x - \tfrac{1}{2}\ln y = 1 + 2p - \tfrac{1}{2}q$ .
$\log_3 \tfrac{54}{2} = \log_3 27 = 3$ .
Domain $x > 2$ . $\log_2\dfrac{x + 1}{x - 2} = 2 \Rightarrow \dfrac{x + 1}{x - 2} = 4 \Rightarrow x + 1 = 4x - 8 \Rightarrow x = 3$ . Valid.
Domain $x > 3$ . $\lg\big(x(x - 3)\big) = 1 \Rightarrow x^2 - 3x = 10 \Rightarrow (x - 5)(x + 2) = 0$ . Reject $x = -2$ ; $x = 5$ .
Domain $x > 0$ . $\ln(x + 2) = \ln(2x^2) \Rightarrow 2x^2 - x - 2 = 0 \Rightarrow x = \dfrac{1 \pm \sqrt{17}}{4}$ . The negative root is rejected, so $x = \dfrac{1 + \sqrt{17}}{4}$ ( $\approx 1.28$ ).
(a) $\ln y = \ln e^2 + \ln x^3 = \ln(e^2x^3)$ , so $y = e^2x^3$ . (b) $\log_2 y + \log_2 x = 3 \Rightarrow xy = 8 \Rightarrow y = \dfrac{8}{x}$ .
$\ln(x - y)^2 = \ln(xy) \Rightarrow (x - y)^2 = xy \Rightarrow x^2 - 2xy + y^2 = xy \Rightarrow x^2 - 3xy + y^2 = 0$ . Divide by $y^2$ and let $t = \tfrac{x}{y}$ : $t^2 - 3t + 1 = 0$ , so $t = \dfrac{3 \pm \sqrt{5}}{2}$ . Since $x > y$ , $t > 1$ , so $\dfrac{x}{y} = \dfrac{3 + \sqrt{5}}{2}$ . (The other root, $\tfrac{3 - \sqrt{5}}{2} \approx 0.38$ , would give $x < y$ , making $\ln(x - y)$ undefined.)

Logarithms are powers

The three laws

Simplifying and combining

Solving equations involving logarithms

Rearranging relationships

Common mistakes

Summary

Practice

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