Add Maths · 6 Logarithmic and exponential functions

6 Logarithmic and exponential functions

Before you start

The laws of indices: $a^{m} \times a^{n} = a^{m + n}$ , $a^{m} \div a^{n} = a^{m - n}$ , ${(a^{m})}^{n} = a^{m n}$ , $a^{0} = 1$ , $a^{- n} = \frac{1}{a^{n}}$ , $a^{\frac{1}{2}} = \sqrt{a}$ .
Solving quadratic equations by factorising.
What an inverse function is: it undoes the original, and its graph is the reflection of the original in the line $y = x$ .
Sketching a graph from its key features: where it meets the axes and what happens for large positive and negative $x$ .

Syllabus objectives covered (3)

Know and use simple properties and graphs of the logarithmic and exponential functions, including ln x and eˣ.
Know and use the laws of logarithms, including change of base of logarithms.
Solve equations of the form aˣ = b.

Part 1 of 8

What a logarithm is

A logarithm answers one question: what power? $\log_{2} 8$ asks 'what power of $2$ gives $8$ ?' The answer is $3$ , because $2^{3} = 8$ .
So every log statement is an index statement written the other way round. Switching between the two forms is the core skill of this whole topic.
$\log_{a} b = c ⟺ a^{c} = b$
Work some out by asking the question. $\log_{3} \frac{1}{9} = - 2$ because $3^{- 2} = \frac{1}{9}$ . $\log_{4} 2 = \frac{1}{2}$ because $4^{\frac{1}{2}} = 2$ .
Know two values on sight: $\log_{a} 1 = 0$ because $a^{0} = 1$ , and $\log_{a} a = 1$ because $a^{1} = a$ .
The base $a$ is positive and not $1$ , and you can only take the log of a positive number, because no power of a positive base is ever zero or negative. So $\log_{2} (- 4)$ does not exist.
Logarithms can have any base. Two have their own names: $\lg x$ means $\log_{10} x$ , and $\ln x$ means $\log_{e} x$ , the natural logarithm, which comes later in this lesson.

Part 2 of 8

The laws of logarithms

You must recall these; they are not on the List of formulas. Each one is an index law in disguise, which is how you rebuild it if you forget.
$\log_{a} (x y) = \log_{a} x + \log_{a} y \log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y \log_{a} (x^{k}) = k \log_{a} x$
Why the first law holds: if $x = a^{m}$ and $y = a^{n}$ , then $x y = a^{m + n}$ , so $\log_{a} (x y) = m + n = \log_{a} x + \log_{a} y$ . Multiplying numbers adds their powers, so it adds their logs.
The power law is the most useful of the three, because it brings an unknown down out of a power: $\log_{a} (x^{k}) = k \log_{a} x$ . That is how logs solve equations like $3^{x} = 5$ .
Two identities come straight from the definition. They say that $\log_{a}$ and 'raise $a$ to the power' undo each other.
$\log_{a} (a^{x}) = x a^{\log_{a} x} = x for x > 0$
Know what the laws do not say. There is no law for $\log_{a} (x + y)$ . And $\frac{\log_{a} x}{\log_{a} y}$ is one number divided by another; it is not $\log_{a} \frac{x}{y}$ .

Part 3 of 8

Writing an expression as a single logarithm

The order that avoids mistakes: first use the power law to move every coefficient up into an index, then combine with the product and quotient laws.
A plain number has to become a log in the same base before it can combine. In base $10$ , $2 = \lg 100$ because $10^{2} = 100$ . In base $5$ , $2 = \log_{5} 25$ .
Write $2 - \lg 4 + 2 \lg x$ as a single logarithm. Power law first, $2 \lg x = \lg x^{2}$ , and replace $2$ by $\lg 100$ .
$\lg 100 - \lg 4 + \lg x^{2}$
Subtracting a log divides, and adding a log multiplies.
$\lg (\frac{100 x^{2}}{4}) = \lg (25 x^{2})$
The same laws turn a log equation into an ordinary one. To solve $\lg (x + 3) + \lg x = 1$ , combine to $\lg (x (x + 3)) = 1$ , then switch to index form: $x (x + 3) = 10^{1}$ .
$x^{2} + 3 x - 10 = 0 \Rightarrow (x + 5) (x - 2) = 0$
That gives $x = 2$ or $x = - 5$ . **Reject $x = - 5$ **: it would need $\lg$ of a negative number, which does not exist. The answer is $x = 2$ only. Always test the solutions of a log equation in the original logs.

Part 4 of 8

Change of base

Logs in different bases cannot be combined directly. The change of base law rewrites a log in any base you choose. You must recall this one too.
$\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$
It comes from the power law. If $\log_{a} b = y$ , then $a^{y} = b$ . Take $\log_{c}$ of both sides: $y \log_{c} a = \log_{c} b$ , so $y = \frac{\log_{c} b}{\log_{c} a}$ .
Choosing $c = b$ gives a special case worth spotting, because it turns a log upside down.
$\log_{a} b = \frac{1}{\log_{b} a}$
So $\frac{1}{\log_{3} e} = \log_{e} 3 = \ln 3$ . Taking the reciprocal of a log swaps its base and the number inside it.
For mixed bases in one equation, rewrite everything in the smallest base. $\log_{4} x = \frac{\log_{2} x}{\log_{2} 4} = \frac{1}{2} \log_{2} x$ . So $\log_{2} x + \log_{4} x = 6$ becomes $\frac{3}{2} \log_{2} x = 6$ , giving $\log_{2} x = 4$ and $x = 2^{4} = 16$ .
The same law evaluates awkward logs by hand: $\log_{4} 8 = \frac{\log_{2} 8}{\log_{2} 4} = \frac{3}{2}$ .

Part 5 of 8

The number e, and ln x and eˣ as inverses

$e$ is a constant, about $2.718$ , and $y = e^{x}$ is the exponential function. On a non-calculator paper you never need its decimal value: answers stay in terms of $e$ and $\ln$ .
$\ln x$ is $\log_{e} x$ , so every law above applies to it unchanged: $\ln (x y) = \ln x + \ln y$ , $\ln x^{k} = k \ln x$ , $\ln 1 = 0$ and $\ln e = 1$ .
The syllabus wants you to understand that $f (x) = e^{x}$ and $g (x) = \ln x$ are each the inverse of the other. In symbols, each one undoes the other.
$\ln (e^{x}) = x for all x, e^{\ln x} = x for x > 0$
That is how you undo either one in an equation. $e^{x} = 5$ gives $x = \ln 5$ . $\ln x = 3$ gives $x = e^{3}$ . $e^{2 x} = 7$ gives $2 x = \ln 7$ , so $x = \frac{1}{2} \ln 7$ .
Because they are inverses, the graph of $y = \ln x$ is the reflection of $y = e^{x}$ in the line $y = x$ . Every point swaps its coordinates: $(0, 1)$ on $y = e^{x}$ becomes $(1, 0)$ on $y = \ln x$ , and $(1, e)$ becomes $(e, 1)$ .
The asymptotes swap too: $y = 0$ for $y = e^{x}$ becomes $x = 0$ for $y = \ln x$ .
Read the key features off that reflection. $y = e^{x}$ is always positive, crosses the $y$ -axis at $(0, 1)$ , and gets closer and closer to the $x$ -axis as $x$ becomes large and negative without ever reaching it. Its asymptote is $y = 0$ .
$y = \ln x$ exists only for $x > 0$ , crosses the $x$ -axis at $(1, 0)$ , and drops steeply alongside the $y$ -axis as $x$ gets close to $0$ . Its asymptote is $x = 0$ . It keeps increasing, slowly, for ever.

Part 6 of 8

Graphs of y = keⁿˣ + a

The exponential graphs in this syllabus have the form $y = k e^{n x} + a$ , with $n$ , $k$ and $a$ integers. Four features describe any of them, and each comes from a short piece of algebra, not from plotting.
Asymptote. $e^{n x}$ gets closer and closer to $0$ on one side: as $x \to - \infty$ if $n > 0$ , or as $x \to + \infty$ if $n < 0$ . On that side $y$ gets closer and closer to $a$ , so the asymptote is $y = a$ . The $+ a$ moves it off the $x$ -axis.
** $y$ -intercept.** Put $x = 0$ . Since $e^{0} = 1$ , $y = k + a$ .
** $x$ -intercept.** Solve $k e^{n x} + a = 0$ , so $e^{n x} = - \frac{a}{k}$ . This has a solution only if $- \frac{a}{k}$ is positive, because $e^{n x}$ is always positive. Otherwise the curve never crosses the $x$ -axis.
Direction. $e^{n x}$ increases if $n > 0$ and decreases if $n < 0$ . A negative $k$ reverses that, and the curve then approaches its asymptote from below instead of from above.
Put it together for $y = 3 e^{2 x} - 6$ . The asymptote is $y = - 6$ , approached as $x \to - \infty$ . The $y$ -intercept is $3 - 6 = - 3$ . The $x$ -intercept comes from $e^{2 x} = 2$ , so $2 x = \ln 2$ and $x = \frac{1}{2} \ln 2$ . With $n = 2$ and $k = 3$ both positive, the curve rises more and more steeply to the right.
The asymptote is $y = a$ , here $y = - 6$ , not the $x$ -axis, and the intercepts are labelled exactly.
On a sketch, label exact values such as $(\frac{1}{2} \ln 2, 0)$ , not decimals, and draw the asymptote as a dashed line with its equation written beside it.

Part 7 of 8

Graphs of y = k ln(ax + b)

The logarithmic graphs in this syllabus have the form $y = k \ln (a x + b)$ , with $k$ , $a$ and $b$ integers. One rule drives everything: $\ln$ only accepts positive inputs, so the graph exists only where $a x + b > 0$ .
Asymptote. The graph drops (or rises, if $k < 0$ ) steeply at the edge of that region, where $a x + b = 0$ . The vertical asymptote is $x = - \frac{b}{a}$ .
** $x$ -intercept.** $\ln$ is zero only at $1$ , so solve $a x + b = 1$ , which gives $x = \frac{1 - b}{a}$ .
** $y$ -intercept.** Put $x = 0$ to get $y = k \ln b$ . This only exists if $x = 0$ is inside the region, which needs $b > 0$ . Otherwise the curve does not meet the $y$ -axis.
For $y = 2 \ln (x + 3)$ : the graph exists for $x > - 3$ and the asymptote is $x = - 3$ . The $x$ -intercept is where $x + 3 = 1$ , which is $(- 2, 0)$ . The $y$ -intercept is $(0, 2 \ln 3)$ .
Nothing exists to the left of the asymptote, because $\ln$ of a negative number is undefined.
If $a$ is negative, the region is on the left instead. $y = \ln (4 - 2 x)$ exists for $x < 2$ , its asymptote $x = 2$ is on the right, and the curve decreases as $x$ increases.

Part 8 of 8

Solving aˣ = b

First look for a common base. $8^{x} = 32$ is $2^{3 x} = 2^{5}$ , so $3 x = 5$ and $x = \frac{5}{3}$ . When both sides are powers of the same number, compare the powers.
With no common base, take logs of both sides (any base) and use the power law to bring $x$ down. From $3^{x} = 5$ , $x \lg 3 = \lg 5$ .
$x = \frac{\lg 5}{\lg 3} = \log_{3} 5$
On a non-calculator paper that is the answer. $\frac{\ln 5}{\ln 3}$ , $\frac{\lg 5}{\lg 3}$ and $\log_{3} 5$ are equal, exact forms. Do not 'simplify' to $\ln \frac{5}{3}$ : that is a different number. As a sanity check, $3 < 5 < 9$ , so $\log_{3} 5$ lies between $1$ and $2$ .
With $x$ in both powers, take logs, expand, collect the $x$ terms and factorise. From $2^{x + 1} = 3^{x}$ : $(x + 1) \ln 2 = x \ln 3$ , so $\ln 2 = x (\ln 3 - \ln 2)$ .
$x = \frac{\ln 2}{\ln 3 - \ln 2} = \frac{\ln 2}{\ln \frac{3}{2}}$
Some equations hide a quadratic. In $4^{x} - 6 (2^{x}) + 8 = 0$ , notice that $4^{x} = {(2^{x})}^{2}$ . Let $u = 2^{x}$ : $u^{2} - 6 u + 8 = 0$ , so $u = 2$ or $u = 4$ . Each becomes a simple $a^{x} = b$ : $2^{x} = 2$ gives $x = 1$ , and $2^{x} = 4$ gives $x = 2$ .
The trap in a hidden quadratic: $a^{x}$ is always positive. For $e^{2 x} - 2 e^{x} - 8 = 0$ , $u = e^{x}$ gives $(u - 4) (u + 2) = 0$ . Keep $e^{x} = 4$ , so $x = \ln 4 = 2 \ln 2$ . Reject $e^{x} = - 2$ and say why: $e^{x} > 0$ for every $x$ .

Worked examples

Example 1. Solve $\log_{3} (2 x + 1) - \log_{3} (x - 1) = 2$ .

Show the working

Combine the two logs with the quotient law, so there is a single log to undo.
$\log_{3} (\frac{2 x + 1}{x - 1}) = 2$
Switch to index form: the base $3$ , raised to the right-hand side, equals what is inside the log.
$\frac{2 x + 1}{x - 1} = 3^{2} = 9$
Multiply both sides by $(x - 1)$ : $2 x + 1 = 9 x - 9$ , so $10 = 7 x$ and $x = \frac{10}{7}$ .
Check the logs exist: $x - 1 = \frac{3}{7} > 0$ and $2 x + 1 = \frac{27}{7} > 0$ . Their ratio is $\frac{27}{7} \div \frac{3}{7} = 9 = 3^{2}$ , so the equation holds.

Answer: $x = \frac{10}{7}$ .

Example 2. Solve $\log_{2} x = 4 \log_{x} 2$ .

Show the working

The bases differ, and one log is the other turned upside down: by change of base, $\log_{x} 2 = \frac{1}{\log_{2} x}$ .
Let $u = \log_{2} x$ so the structure is easy to see.
$u = \frac{4}{u} \Rightarrow u^{2} = 4 \Rightarrow u = 2 or u = - 2$
Keep both. A log can be negative; it is the number inside a log that must be positive. Switch each to index form: $\log_{2} x = 2$ gives $x = 4$ , and $\log_{2} x = - 2$ gives $x = 2^{- 2} = \frac{1}{4}$ .
Check $x = \frac{1}{4}$ . The left side is $- 2$ . On the right, $\log_{1 / 4} 2 = - \frac{1}{2}$ because ${(\frac{1}{4})}^{- \frac{1}{2}} = 4^{\frac{1}{2}} = 2$ , and $4 \times (- \frac{1}{2}) = - 2$ . Both solutions stand.

Answer: $x = 4$ or $x = \frac{1}{4}$ .

Example 3. Solve $5^{2 x - 1} = 3^{x}$ , giving $x$ exactly.

Show the working

There is no common base, so take natural logs of both sides and bring the powers down. Keep the bracket round $2 x - 1$ .
$(2 x - 1) \ln 5 = x \ln 3$
Expand, then collect the $x$ terms on one side.
$2 x \ln 5 - x \ln 3 = \ln 5$
Factorise out $x$ . This is the step that isolates it.
$x (2 \ln 5 - \ln 3) = \ln 5$
Tidy the bracket into a single log: $2 \ln 5 - \ln 3 = \ln 25 - \ln 3 = \ln \frac{25}{3}$ .
$x = \frac{\ln 5}{\ln \frac{25}{3}}$
A sanity check with no calculator: at $x = 0$ the left side is $\frac{1}{5}$ and the right side is $1$ ; at $x = 1$ they are $5$ and $3$ . The sides swap order, so the solution lies between $0$ and $1$ . Since $5 < \frac{25}{3}$ , the fraction $\frac{\ln 5}{\ln \frac{25}{3}}$ is less than $1$ , which fits.

Answer: $x = \frac{\ln 5}{\ln \frac{25}{3}}$ , or equivalently $\frac{\ln 5}{2 \ln 5 - \ln 3}$ .

Example 4. A curve has equation $y = 4 - 2 e^{3 x}$ . State the equation of its asymptote, find where it meets the axes, and describe its shape.

Show the working

Match it to $y = k e^{n x} + a$ : $k = - 2$ , $n = 3$ and $a = 4$ .
Asymptote: $n > 0$ , so $e^{3 x} \to 0$ as $x \to - \infty$ , and then $y \to 4$ . The asymptote is $y = 4$ .
$y$ -intercept: $x = 0$ gives $y = 4 - 2 (1) = 2$ , so the point is $(0, 2)$ .
$x$ -intercept: $4 - 2 e^{3 x} = 0$ gives $e^{3 x} = 2$ , so $3 x = \ln 2$ and $x = \frac{1}{3} \ln 2$ . It exists because $2$ is positive.
Shape: $e^{3 x}$ increases, and $k = - 2$ is negative, so $y$ decreases as $x$ increases. The curve starts just below $y = 4$ on the far left, passes through $(0, 2)$ and $(\frac{1}{3} \ln 2, 0)$ , then falls more and more steeply.
Why 'just below'? $2 e^{3 x}$ is always positive, so $y = 4 - 2 e^{3 x}$ is always less than $4$ . The curve never reaches its asymptote.

Answer: Asymptote $y = 4$ . It meets the $y$ -axis at $(0, 2)$ and the $x$ -axis at $(\frac{1}{3} \ln 2, 0)$ . It is decreasing, approaching $y = 4$ from below as $x \to - \infty$ .

Where people go wrong

Mistake: Writing $\lg (x + 3) = \lg x + \lg 3$ .

Instead: There is no law for the log of a sum. It is $\lg x + \lg 3$ that equals $\lg 3 x$ .

The product law comes from adding powers when numbers are multiplied. Nothing like that happens when numbers are added.

Mistake: 'Simplifying' $\frac{\ln 5}{\ln 3}$ to $\ln \frac{5}{3}$ or to $\ln 5 - \ln 3$ .

Instead: Leave $\frac{\ln 5}{\ln 3}$ , or write it as $\log_{3} 5$ .

$\ln \frac{5}{3}$ is the log of a quotient. $\frac{\ln 5}{\ln 3}$ is a quotient of logs, which by change of base is $\log_{3} 5$ , a different number.

Mistake: Writing $2 \lg x$ as $\lg 2 x$ .

Instead: $2 \lg x = \lg x^{2}$ .

The power law moves the coefficient up into a power, not into a product.

Mistake: Giving the asymptote of $y = 3 e^{2 x} - 6$ as $y = 0$ .

Instead: The asymptote is $y = - 6$ .

As $e^{2 x} \to 0$ , $y \to 0 - 6$ . The constant $a$ moves the asymptote along with the rest of the graph.

Mistake: Keeping a solution that puts a negative number inside a log, or accepting $e^{x} = - 2$ .

Instead: Test every solution in the original equation. Reject values that need the log of a number that is not positive, or a power equal to a negative number.

$a^{x} > 0$ for every $x$ , and a log accepts only positive inputs. The algebra can produce values that were never possible.

Check yourself

Find the exact value of (a) $\log_{2} 32$ (b) $\log_{9} 3$ (c) $\lg 0.01$ (d) $\ln e^{4}$ .
Answer
(a) $5$ (b) $\frac{1}{2}$ , since $9^{\frac{1}{2}} = 3$ (c) $- 2$ (d) $4$ .
Write $\frac{1}{\log_{7} e}$ as a natural logarithm.
Answer
$\ln 7$ , because $\log_{a} b = \frac{1}{\log_{b} a}$ .
Write $3 \lg 2 + \lg x - 1$ as a single logarithm.
Answer
$\lg 8 + \lg x - \lg 10 = \lg \frac{8 x}{10} = \lg \frac{4 x}{5}$ .
Solve $8^{x} = 4^{x + 1}$ .
Answer
Write both as powers of $2$ : $2^{3 x} = 2^{2 x + 2}$ , so $3 x = 2 x + 2$ and $x = 2$ .
Solve $5^{x} = 12$ , giving your answer exactly, and state which two whole numbers it lies between.
Answer
$x = \frac{\lg 12}{\lg 5} = \log_{5} 12$ . Since $5 < 12 < 25$ , it lies between $1$ and $2$ .
Solve $2 e^{2 x} - 7 e^{x} + 3 = 0$ .
Answer
Let $u = e^{x}$ : $2 u^{2} - 7 u + 3 = (2 u - 1) (u - 3) = 0$ , so $e^{x} = \frac{1}{2}$ or $e^{x} = 3$ . Both are positive, so $x = \ln \frac{1}{2} = - \ln 2$ or $x = \ln 3$ .
For $y = 5 - 3 e^{- 2 x}$ , state the equation of the asymptote and the $y$ -intercept.
Answer
Here $n = - 2 < 0$ , so $e^{- 2 x} \to 0$ as $x \to + \infty$ , and the asymptote is $y = 5$ . The $y$ -intercept is $5 - 3 = 2$ .
State the equation of the asymptote of $y = \ln (2 x - 6)$ and find where the curve meets the $x$ -axis.
Answer
It needs $2 x - 6 > 0$ , so the asymptote is $x = 3$ . Setting $2 x - 6 = 1$ gives $x = \frac{7}{2}$ , so it meets the $x$ -axis at $(\frac{7}{2}, 0)$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Write $1 + 2 \lg x - \lg 5$ as a single base 10 logarithm, simplifying your answer. 3 marks No calculator

Mark scheme

B1 $1 = \lg 10$ soi
M1 power law and at least one of the product or quotient laws applied correctly
A1 $\lg (2 x^{2})$ cao

(b) Use the change of base law to solve $\log_{3} x + \log_{9} x = 3$ . 3 marks No calculator

Mark scheme

M1 $\log_{9} x = \frac{\log_{3} x}{\log_{3} 9}$ oe
A1 $\frac{3}{2} \log_{3} x = 3$ oe
A1 $x = 9$

Question 2.

(a) Solve $3^{x + 2} = 7^{x}$ , giving your answer in the form $\frac{\ln p}{\ln q}$ , where $p$ and $q$ are rational numbers. 4 marks No calculator

Mark scheme

M1 takes logs of both sides and brings the powers down: $(x + 2) \ln 3 = x \ln 7$ oe
M1 collects the $x$ terms and factorises: $x (\ln 7 - \ln 3) = 2 \ln 3$ oe
A1 $x = \frac{2 \ln 3}{\ln 7 - \ln 3}$ oe
A1 $x = \frac{\ln 9}{\ln \frac{7}{3}}$ cao

(b) Solve $9^{x} - 3^{x + 1} - 18 = 0$ , giving your answer exactly. 4 marks No calculator

Mark scheme

B1 $9^{x} = {(3^{x})}^{2}$ and $3^{x + 1} = 3 \times 3^{x}$ used, e.g. $u^{2} - 3 u - 18 = 0$
M1 solves their three-term quadratic in $u$
A1 $3^{x} = 6$ , with $3^{x} = - 3$ rejected because $3^{x} > 0$
A1 $x = \frac{\ln 6}{\ln 3}$ oe, e.g. $\log_{3} 6$ or $1 + \log_{3} 2$

Question 3. The curve $C$ has equation $y = 3 \ln (2 x - 4)$ .

(a) State the equation of the asymptote of $C$ . 1 mark No calculator

Mark scheme

B1 $x = 2$

(b) Sketch $C$ , showing the asymptote and the coordinates of any point where $C$ meets the axes. 3 marks No calculator

Mark scheme

B1 increasing logarithmic shape, entirely to the right of $x = 2$
B1 asymptote $x = 2$ drawn and labelled, with the curve falling steeply towards it
B1 $(\frac{5}{2}, 0)$ labelled, and no intersection with the $y$ -axis

(c) Solve $3 \ln (2 x - 4) = 6$ , giving your answer exactly. 2 marks No calculator

Mark scheme

M1 $\ln (2 x - 4) = 2$ and $2 x - 4 = e^{2}$
A1 $x = \frac{e^{2} + 4}{2}$ oe

How this is examined

Solving an exponential equation exactly, by taking logs or by spotting a hidden quadratic: a four- or five-mark Solve, sometimes asking for the answer in a stated form such as $\frac{\ln p}{\ln q}$ Logs taken of both sides with the powers brought down in brackets (M), the $x$ terms collected and factorised (M), and an exact final answer. For a hidden quadratic: $u = a^{x}$ stated, both values of $u$ found, and the negative one rejected with the reason $a^{x} > 0$ .
Combining logs and solving a log equation: a Solve with two or three log terms in one base, or a Write as a single logarithm The power law used before the product and quotient laws, plain numbers written as logs of the same base (e.g. $1 = \lg 10$ ), the single log converted to index form, and any solution that puts a non-positive number inside a log rejected.
Key features of $y = k e^{n x} + a$ or $y = k \ln (a x + b)$ : a State the asymptote, then a Sketch worth two or three marks, sometimes with intercepts to Find The correct shape on the correct side of a dashed asymptote labelled with its equation ( $y = a$ or $x = - \frac{b}{a}$ ), with intercepts labelled exactly, e.g. $(\frac{1}{2} \ln 2, 0)$ .
Changing base to handle mixed bases: a Solve such as $\log_{3} x + \log_{9} x = k$ , or a Use the change of base law The change of base written out explicitly, e.g. $\log_{9} x = \frac{\log_{3} x}{\log_{3} 9} = \frac{1}{2} \log_{3} x$ , before the terms are combined.
Undoing $e$ with $\ln$ , and $\ln$ with $e$ : a one- or two-mark step inside a longer question $e^{k x} = c$ rewritten as $k x = \ln c$ , or $\ln (\dots) = c$ rewritten as $(\dots) = e^{c}$ , with the answer left exact.

Words to know

Logarithm: $\log_{a} b$ is the power to which $a$ must be raised to give $b$ .
lg: Shorthand for $\log_{10}$ .
Natural logarithm: $\ln x = \log_{e} x$ , the inverse function of $e^{x}$ .
e: A constant, about $2.718$ , which is the base of the exponential function $e^{x}$ and of $\ln$ .
Asymptote: A line that a curve gets closer and closer to without reaching, stated as an equation such as $y = - 6$ or $x = - 3$ .
Change of base: $\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$ , which rewrites a log in any base.
Exponential equation: An equation with the unknown in a power, such as $a^{x} = b$ .

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