Logarithmic and Exponential Functions

Syllabus coverage

Understand the relationship between logarithms and indices and use the laws of logarithms, excluding change of base
Understand the definition and properties of $e^{x}$ and $\ln x$ , including their relationship as inverse functions and their graphs, and know the graph of $y = e^{k x}$ for positive and negative values of $k$
Use logarithms to solve equations and inequalities in which the unknown appears in an index
Use logarithms to transform a relationship to linear form, and determine unknown constants from the gradient and the intercept of the resulting straight line

Part of the AS & A Level Maths syllabus 9709.

The two shapes every exponential curve has. For

k > 0

the graph of

y = e^{k x}

climbs away from the x-axis; for

k < 0

it falls towards it. The straight line shows how differently exponential change behaves from a fixed change.

A logarithm is an index written the other way round

The definition is the whole topic: $\log_{a} x = n$ means exactly the same thing as $a^{n} = x$ . The number $a$ is the base, and it must be positive and not equal to 1; $x$ must be positive, because no power of a positive base is negative or zero. So $\log_{2} 8 = 3$ because $2^{3} = 8$ , and $\log_{5} \frac{1}{25} = - 2$ because $5^{- 2} = \frac{1}{25}$ .

Two bases are used throughout Paper 3. Base 10 gives the common logarithm, written $\log x$ with no base shown, and it is the one on a calculator button. Base $e$ gives the natural logarithm, written $\ln x$ , and it is the one the rest of the course is built on. Nothing else needs a base of its own.

Four consequences of the definition are used constantly and are worth knowing cold: $\log_{a} 1 = 0$ because $a^{0} = 1$ ; $\log_{a} a = 1$ ; $\log_{a} a^{k} = k$ because the two operations cancel; and $a^{\log_{a} x} = x$ . The last two are the same statement seen from each side, and they are what make logarithms and exponentials inverse functions.

Logarithms turn multiplication into addition, which is why they were invented and why they still solve equations that nothing else can touch. A question that says 'solve', where the unknown is in the air, is almost always a question about taking logarithms of both sides.

$\log_{a} x = n \Leftrightarrow a^{n} = x$ - learn this as one two-way statement
$\log_{a} 1 = 0$ and $\log_{a} a = 1$ for every valid base
$\log_{a} a^{k} = k$ , so $\log_{2} 2^{7} = 7$ without any calculation
You cannot take the logarithm of 0 or of a negative number
$\log_{3} \frac{1}{9} = - 2$ , not $- \frac{1}{2}$ : the answer is the index, not the value of the fraction

The answer to a logarithm is an index. If a question asks for

\log_{2} 32

the answer is 5, and writing

2^{5}

is answering a different question - it is the working, not the answer.

The laws of logarithms

Three laws cover everything Paper 3 asks. The product law: $\log_{a} x y = \log_{a} x + \log_{a} y$ . The quotient law: $\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y$ . The power law: $\log_{a} x^{k} = k \log_{a} x$ . All three come from the index laws, and all three need the same base on every logarithm before they can be used.

They are used in both directions. Combining several logarithms into one is what lets you solve an equation: $\log_{2} x + \log_{2} (x + 2) = 3$ becomes $\log_{2} [x (x + 2)] = 3$ , and then the definition turns it into $x (x + 2) = 2^{3} = 8$ , so $x^{2} + 2 x - 8 = 0$ , giving $x = 2$ or $x = - 4$ . The solution $x = - 4$ must be rejected, because $\log_{2} (- 4)$ does not exist - a logarithm question that produces two roots almost always wants one of them thrown away with a reason.

Splitting a single logarithm into several is what the later topics need. The power law is how a logarithm of a power is written, and it is the step that makes an equation with the unknown in the index solvable at all. You will use it again in differentiation and integration, where $\ln x$ and $e^{x}$ are the functions that behave most simply.

Change of base is deliberately excluded at 9709, so you will never need to convert a logarithm from one base to another by formula. If an equation involves base 10 and base $e$ together, take logarithms in whichever base you like on both sides and carry the base through to the end.

Law	Statement	Example
Product	$\log_{a} x y = \log_{a} x + \log_{a} y$	$\log_{2} 4 + \log_{2} 8 = \log_{2} 32 = 5$
Quotient	$\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y$	$\log_{3} 18 - \log_{3} 2 = \log_{3} 9 = 2$
Power	$\log_{a} x^{k} = k \log_{a} x$	$\log_{5} 25^{x} = 2 x$
Special values	$\log_{a} 1 = 0$ , $\log_{a} a = 1$	$\ln 1 = 0$ , $\ln e = 1$
Cancelling	$a^{\log_{a} x} = x$ , $\log_{a} a^{k} = k$	$e^{\ln 7} = 7$ , $\ln e^{4} = 4$

\log_{a} (x + y)

is not

\log_{a} x + \log_{a} y

. The product law is about the logarithm of a product; there is no law at all for the logarithm of a sum, and inventing one is the single commonest error in this topic.

The exponential function e^x and the natural logarithm ln x

$e$ is a number, approximately 2.718281828, chosen because $y = e^{x}$ is the one exponential curve whose gradient at every point equals its height. $e^{x}$ is called the exponential function, and $\ln x$ means $\log_{e} x$ , the natural logarithm.

Because they share a base, $e^{x}$ and $\ln x$ undo each other: $\ln (e^{x}) = x$ for every real $x$ , and $e^{\ln x} = x$ for every $x > 0$ . That second restriction matters. $e^{\ln (- 3)}$ is meaningless, because $\ln (- 3)$ does not exist, so a solution that requires it must be rejected.

The two functions are therefore inverse functions, and this is the relationship the syllabus asks you to understand rather than merely quote. It is what lets you solve $\ln (2 x - 1) = 3$ : exponentiate both sides to get $2 x - 1 = e^{3}$ , so $x = \frac{e^{3} + 1}{2} = 10.5$ to 3 significant figures. In the other direction, $e^{2 x} = 7$ gives $2 x = \ln 7$ , so $x = \frac{\ln 7}{2} = 0.973$ .

Two facts about the ranges are worth memorising because they decide how many solutions an equation has. $e^{x}$ is positive for every $x$ , so $e^{x} = - 4$ has no solutions. And $\ln x$ is defined only for $x > 0$ , so $\ln x = - 4$ does have a solution, namely $x = e^{- 4}$ .

Statement	Equivalent statement	Restriction
$e^{0} = 1$	$\ln 1 = 0$	none
$e^{1} = e$	$\ln e = 1$	none
$\ln (e^{x}) = x$	$e^{\ln x} = x$	$x > 0$ in the second form only
$e^{x} > 0$ always	$\ln x$ is undefined for $x \leq 0$	the domain of $\ln x$ is $x > 0$

e^{x}

never reaches zero and never goes negative, so the x-axis is an asymptote of its graph.

\ln x

is the mirror image, so the y-axis is an asymptote of that one.

The graphs of y = e^x, y = ln x and y = e^kx

The graph of $y = e^{x}$ passes through $(0, 1)$ , climbs steeply to the right, and approaches the x-axis as $x$ becomes very negative without ever touching it. Its range is $y > 0$ and it is increasing everywhere. The graph of $y = \ln x$ passes through $(1, 0)$ , climbs more and more slowly, and approaches the y-axis as $x$ approaches zero from above. Its domain is $x > 0$ and its range is every real number.

Because the functions are inverses, one graph is the reflection of the other in the line $y = x$ . The point $(0, 1)$ on $y = e^{x}$ reflects to $(1, 0)$ on $y = \ln x$ , and the horizontal asymptote of one becomes the vertical asymptote of the other. If a question asks you to sketch both, the line $y = x$ is the mirror, and drawing it makes the sketch easy to justify.

The syllabus asks specifically for $y = e^{k x}$ for both positive and negative $k$ . For $k > 0$ the curve grows as $x$ increases; for $k < 0$ it decays towards the x-axis. Every one of them passes through $(0, 1)$ , because $e^{0} = 1$ whatever $k$ is, and the bigger $| k |$ is, the faster the curve rises or falls. The case $k = - 1$ is worth knowing separately: $y = e^{- x}$ is the reflection of $y = e^{x}$ in the y-axis.

One algebraic fact ties the family together: $e^{k x} = {(e^{k})}^{x}$ , so a curve of the form $y = e^{k x}$ is the same kind of thing as $y = a^{x}$ with $a = e^{k}$ . That is why every exponential model in the paper, whether written with a base of 2 or a base of $e$ , has the same shape and the same two possibilities - growth or decay.

Curve	Point it always passes through	Asymptote	Behaviour
$y = e^{x}$	$(0, 1)$	the x-axis, as $x \to - \infty$	increasing, range $y > 0$
$y = e^{- x}$	$(0, 1)$	the x-axis, as $x \to + \infty$	decreasing, range $y > 0$
$y = e^{k x}$ , $k > 0$	$(0, 1)$	the x-axis on the left	growth; steeper for larger $k$
$y = e^{k x}$ , $k < 0$	$(0, 1)$	the x-axis on the right	decay; faster for more negative $k$
$y = \ln x$	$(1, 0)$	the y-axis, as $x \to 0^{+}$	increasing, domain $x > 0$

Every exponential curve passes through

(0, 1)

and every logarithmic curve passes through

(1, 0)

, because

e^{0} = 1

and

\ln 1 = 0

. A sketch that misses these points has missed the definition.

Solving equations in which the unknown is in the index

When the unknown is in the air, take logarithms of both sides. That step is valid because $\ln$ is an increasing function, so it preserves equality, and it works for any base - $\ln$ and $\log_{10}$ are both acceptable, and the exact answer may be written with either. For $2^{x + 1} = 5$ , taking $\ln$ gives $(x + 1) \ln 2 = \ln 5$ , so $x + 1 = \frac{\ln 5}{\ln 2}$ and $x = 1.322$ to 3 decimal places.

When the unknown appears in two indices, use the power law to bring both down, then collect the $x$ terms. For $3^{2 x - 1} = 4^{x + 1}$ : taking logarithms gives $(2 x - 1) \ln 3 = (x + 1) \ln 4$ ; expanding gives $2 x \ln 3 - \ln 3 = x \ln 4 + \ln 4$ ; collecting gives $x (2 \ln 3 - \ln 4) = \ln 3 + \ln 4$ , so $x = \frac{\ln 3 + \ln 4}{2 \ln 3 - \ln 4} = 3.06$ to 3 significant figures. The whole method is one line of expansion and one line of collecting.

Inequalities work the same way with one extra rule: dividing by a negative number reverses the inequality, and a logarithm can be negative when its base is less than 1 or its argument is less than 1. In $2^{1 - x} > 5$ , taking logarithms gives $(1 - x) \ln 2 > \ln 5$ , so $- x \ln 2 > \ln 5 - \ln 2$ , and multiplying by $- 1$ reverses the sign to give $x \ln 2 < \ln 2 - \ln 5$ . Since $\ln 2 > 0$ the division does not reverse anything, and $x < \frac{\ln 2 - \ln 5}{\ln 2} = - 1.32$ to 3 significant figures.

Two smaller habits save marks. State which logarithm you are taking, because that is a method mark in its own right. And keep the exact fraction until the final line: $\frac{\ln 5}{\ln 2}$ rounded to 1.32 and then reused in a later part can cost the accuracy mark, while the exact form never can.

Write 'take $\ln$ of both sides' before you use the power law
$a^{f (x)} = b$ becomes $f (x) \ln a = \ln b$ in one step
With the unknown in two indices, expand both sides and collect the $x$ terms before dividing
Dividing by a negative logarithm reverses an inequality - check the sign of the log before you divide
${0.5}^{x} = 8$ gives $x = \frac{\ln 8}{\ln 0.5} = - 3$ exactly, because $8 = 2^{3}$ and $\ln 0.5 = - \ln 2$

An equation such as

e^{x} = - 2

has no solutions, because

e^{x}

is always positive. Spotting that before you take logarithms saves you from writing

\ln (- 2)

, which does not exist.

Reducing a relationship to linear form

Experimental data often obey a power law $y = k x^{n}$ or an exponential law $y = k a^{x}$ , and neither can be checked by plotting $y$ against $x$ , because both produce curves. Taking logarithms turns each of them into the equation of a straight line, and the constants can then be read off the graph or worked out from two pairs of values.

For $y = k x^{n}$ , take logarithms of both sides and use the product and power laws: $\ln y = \ln k + n \ln x$ . Plotted with $\ln y$ up the vertical axis and $\ln x$ along the horizontal axis, this is a straight line of gradient $n$ and vertical intercept $\ln k$ . The gradient gives the power directly, and the intercept has to be exponentiated to give $k$ .

For $y = k a^{x}$ , the same step gives $\ln y = \ln k + x \ln a$ , which is a straight line when $\ln y$ is plotted against $x$ itself. Here the gradient is $\ln a$ and the vertical intercept is again $\ln k$ . A relationship of the form $y = A e^{b x}$ is the same case with $a = e^{b}$ : the gradient of the straight line is exactly $b$ , and the intercept is $\ln A$ .

Given two pairs of values, the two constants come out of two simultaneous equations. If $y = k x^{n}$ with $y = 96$ when $x = 4$ , and $y = 729$ when $x = 9$ , then $\ln 96 = \ln k + n \ln 4$ and $\ln 729 = \ln k + n \ln 9$ . Subtracting removes $\ln k$ : $\ln 729 - \ln 96 = n (\ln 9 - \ln 4)$ , so $n = \frac{\ln 729 - \ln 96}{\ln 9 - \ln 4} = \frac{2.0274}{0.8109} = 2.5$ . Then $\ln k = \ln 96 - 2.5 \ln 4 = 1.0986$ , so $k = e^{1.0986} = 3$ .

Relationship	Linear form	Plot	Gradient	Intercept
$y = k x^{n}$	$\ln y = \ln k + n \ln x$	$\ln y$ against $\ln x$	$n$	$\ln k$ , so $k = e^{intercept}$
$y = k a^{x}$	$\ln y = \ln k + x \ln a$	$\ln y$ against $x$	$\ln a$ , so $a = e^{gradient}$	$\ln k$
$y = A e^{b x}$	$\ln y = \ln A + b x$	$\ln y$ against $x$	$b$	$\ln A$
$y = A b^{- x}$	$\ln y = \ln A - x \ln b$	$\ln y$ against $x$	$- \ln b$	$\ln A$

The intercept is

\ln k

, not

k

. An answer that reads a graph intercept of 1.0986 and states

k = 1.0986

has lost the final mark;

k = e^{1.0986} = 3

is the answer.

How a Paper 3 logarithms question is built

There are four shapes and each has a routine. Evaluating a logarithm is a recall question - write the index. Solving an equation with the unknown in the index is 'take logarithms, apply the power law, collect'. Solving an equation with logarithms on both sides is 'combine with the laws, use the definition, then reject any invalid root'. Transforming a relationship to linear form is 'take logarithms, state what is plotted, name the gradient and the intercept, then convert the intercept back'.

The marks sit in the statements rather than in the arithmetic. Writing 'take $\ln$ of both sides', naming the gradient as $\ln a$ , and saying why a root is rejected are each worth a mark that a calculator cannot supply. A bare decimal with no line of working rarely scores more than the final accuracy mark.

Rewrite using the definition before doing anything clever - most questions are shorter than they look
Reject any root that makes an argument of a logarithm zero or negative, and say why in one clause
Give the exact answer and the rounded one when the question asks for a stated accuracy
In a linear-form question, state the axes you are plotting on: $\ln y$ against $\ln x$ and $\ln y$ against $x$ are different graphs
Check a solution by substituting it back into the original equation - index equations are easy to check numerically

Key terms you must be able to define

Logarithm: The index to which a base must be raised to give a number: $\log_{a} x = n$ means $a^{n} = x$ .

Base (of a logarithm): The number being raised to a power, written as a subscript in $\log_{a} x$ . It must be positive and not equal to 1.

Argument of a logarithm: The number the logarithm is taken of. It must be positive: $\log_{a} x$ requires $x > 0$ .

Common logarithm: A logarithm to base 10, written $\log x$ with the base left out. It is the one on the calculator button.

Natural logarithm: A logarithm to base $e$ , written $\ln x$ . The one Paper 3 uses most.

e: The irrational number 2.718281828..., the base for which $y = e^{x}$ has gradient equal to its own height.

Exponential function: $e^{x}$ , always positive, increasing everywhere, with the x-axis as an asymptote.

Inverse functions: Two functions that undo each other. $e^{x}$ and $\ln x$ are inverses, so $\ln (e^{x}) = x$ and $e^{\ln x} = x$ for $x > 0$ .

Asymptote: A line a curve approaches but never reaches. $y = e^{x}$ has the x-axis as an asymptote; $y = \ln x$ has the y-axis.

Product law: $\log_{a} x y = \log_{a} x + \log_{a} y$ . It turns a product into a sum.

Quotient law: $\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y$ . It turns a quotient into a difference.

Power law: $\log_{a} x^{k} = k \log_{a} x$ . It brings an index down to the front, which is how index equations are solved.

Exponential equation: An equation with the unknown in an index, such as $2^{x + 1} = 5$ . Solved by taking logarithms of both sides.

Exponential growth and decay: The behaviour of $y = e^{k x}$ : growth when $k > 0$ , decay towards the x-axis when $k < 0$ .

Linear form: A relationship rewritten so that it plots as a straight line, such as $\ln y = \ln k + n \ln x$ .

Logarithmic transformation: Taking logarithms of a relationship to turn a power law or an exponential law into a straight line, so that the gradient and intercept give the constants.

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

Evaluating a logarithm from the definition

$\log_{2} 32$ asks for the power of 2 that gives 32. Since $2^{5} = 32$ , the answer is 5. $\log_{3} \frac{1}{9}$ asks for the power of 3 that gives $\frac{1}{9}$ , and $3^{- 2} = \frac{1}{9}$ , so the answer is $- 2$ . No calculator is needed for either, and the whole question is worth two marks.

An index equation with one unknown index

Solve $5^{x} = 12$ . Take logarithms of both sides: $x \ln 5 = \ln 12$ , so $x = \frac{\ln 12}{\ln 5} = 1.544$ to 3 decimal places. The exact form $\frac{\ln 12}{\ln 5}$ is the answer to the question; the decimal is what a stated accuracy asks for. Check roughly: $5^{1.5} = 11.18$ , so an answer near 1.54 is sensible.

Exact form first, decimal second

Solve $3^{2 x} = 20$ . Take logarithms: $2 x \ln 3 = \ln 20$ , so $x = \frac{\ln 20}{2 \ln 3} = 1.363$ to 3 significant figures. Notice that the 2 stays with the $x$ rather than disappearing into the fraction: the power law gives $2 x \ln 3$ , and dividing by $2 \ln 3$ is the last step.

Finding the constants of a power law

Data fit $y = k x^{n}$ , with $y = 96$ when $x = 4$ and $y = 729$ when $x = 9$ . Taking logarithms gives $\ln y = \ln k + n \ln x$ . Subtracting the two equations: $n = \frac{\ln 729 - \ln 96}{\ln 9 - \ln 4} = \frac{2.0274}{0.8109} = 2.5$ . Then $\ln k = \ln 96 - 2.5 \ln 4 = 1.0986$ , so $k = e^{1.0986} = 3$ . Check: $3 \times 9^{2.5} = 3 \times 243 = 729$ , as given.

An exponential decay model

The mass $m$ grams of a sample after $t$ days is modelled by $m = 40 e^{- 0.05 t}$ . Find the time at which the mass is 10 grams. Put $40 e^{- 0.05 t} = 10$ , so $e^{- 0.05 t} = 0.25$ . Take logarithms: $- 0.05 t = \ln 0.25$ , so $t = \frac{\ln 0.25}{- 0.05} = 27.7$ days to 3 significant figures. The negative sign in the index is handled by the logarithm, not by ignoring it.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What does

\log_{a} x = n

mean?

a^{n} = x

. A logarithm is an index:

\log_{2} 32 = 5

because

2^{5} = 32

Q2What is

\log_{a} 1

0, for every valid base, because

a^{0} = 1

. So

\ln 1 = 0

Q3What is

\log_{a} a

1, because

a^{1} = a

. So

\ln e = 1

Q4Write down the product law of logarithms.

\log_{a} x y = \log_{a} x + \log_{a} y

. It only applies to a product, never to a sum.

Q5Write down the quotient law of logarithms.

\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y

Q6Write down the power law of logarithms.

\log_{a} x^{k} = k \log_{a} x

. This is the law that frees an unknown from an index.

Q7Why can you not take the logarithm of a negative number?

Because

a^{n}

is positive for every positive base

a

and every

n

, so no index produces a negative result.

Q8What is the relationship between

e^{x}

and

\ln x

They are inverse functions:

\ln (e^{x}) = x

for all

x

, and

e^{\ln x} = x

for

x > 0

Q9Which point does every curve

y = e^{k x}

pass through?

(0, 1)

, because

e^{0} = 1

whatever

k

is.

Q10Which point does

y = \ln x

pass through?

(1, 0)

, because

\ln 1 = 0

Q11What are the asymptotes of

y = e^{x}

and

y = \ln x

y = e^{x}

has the x-axis as an asymptote as

x \to - \infty

;

y = \ln x

has the y-axis as an asymptote as

x \to 0^{+}

Q12What is the shape of

y = e^{k x}

when

k

is negative?

A decreasing decay curve through

(0, 1)

falling towards the x-axis.

y = e^{- x}

is the reflection of

y = e^{x}

in the y-axis.

Q13How do you solve

a^{f (x)} = b

Take logarithms of both sides:

f (x) \ln a = \ln b

, so

f (x) = \frac{\ln b}{\ln a}

. Then solve for

x

Q14What happens to an inequality when you divide by

\ln 0.5

The direction reverses, because

\ln 0.5 = - 0.693

is negative. Check the sign of the logarithm before dividing.

Q15How is

y = k x^{n}

turned into a straight line?

\ln y = \ln k + n \ln x

, so plot

\ln y

against

\ln x

. The gradient is

n

and the intercept is

\ln k

Q16How is

y = k a^{x}

turned into a straight line?

\ln y = \ln k + x \ln a

, so plot

\ln y

against

x

. The gradient is

\ln a

and the intercept is

\ln k

Q17A graph of

\ln y

against

x

has intercept 0.693. What is the constant

k

k = e^{0.693} = 2

. The intercept gives

\ln k

, so it must be exponentiated.

0 of 17 marked as known

Exam technique for this topic

Learn the definition as a two-way statement and use it in the first line of any logarithm question. 'Rewrite $\log_{a} x = n$ as $a^{n} = x$ ' is a method mark, and it turns most equations into ordinary algebra.
When the unknown is in an index, write 'take $\ln$ of both sides' on its own line before applying the power law. The method mark is awarded for the logarithm, not for the eventual answer.
Reject invalid roots explicitly. In $\log_{2} x + \log_{2} (x + 2) = 3$ the quadratic gives $x = 2$ and $x = - 4$ , and the answer is only complete when you have said that $x = - 4$ is rejected because $\log_{2} (- 4)$ is undefined.
In a linear-form question, state the axes and both constants in words: 'plotting $\ln y$ against $\ln x$ gives a straight line of gradient $n$ and intercept $\ln k$ '. Then convert the intercept with $k = e^{intercept}$ - the graph gives you the logarithm of $k$ , never $k$ itself.
Give both the exact form and the rounded value when an accuracy is specified. $\frac{\ln 3 + \ln 4}{2 \ln 3 - \ln 4}$ earns the exactness mark, and 3.06 earns the accuracy mark; writing only 3.06 leaves the first of those on the table.

Why students lose marks here

A student wrote

\log_{2} 6 = \log_{2} 2 + \log_{2} 4

, so the product law lets you split a logarithm of a sum.

Why it loses marksThe product law is

\log_{a} x y = \log_{a} x + \log_{a} y

, which is about a product inside the logarithm. There is no law at all for the logarithm of a sum, and

\log_{2} 6

is not

1 + 2

The answer that scores

\log_{2} 6

stays as it is.

\log_{2} 2 + \log_{2} 4 = 1 + 2 = 3

, which is

\log_{2} 8

, not

\log_{2} 6

A student wroteFrom

\log_{2} x + \log_{2} (x + 2) = 3

, writing

x (x + 2) = 3

Why it loses marksThe 3 is the value of the logarithm, so the definition gives

x (x + 2) = 2^{3}

. Losing the base is the commonest single error in this topic.

The answer that scoresCombine first:

\log_{2} [x (x + 2)] = 3

, so

x (x + 2) = 8

, giving

x^{2} + 2 x - 8 = 0

x = 2

x = - 4

. The root

x = - 4

is rejected, so

x = 2

A student wroteDividing by

\ln 0.5

to solve

{0.5}^{x} > 3

and writing

x > \frac{\ln 3}{\ln 0.5}

Why it loses marks

\ln 0.5

is negative, so dividing by it reverses the direction of the inequality. The arithmetic is right and the inequality sign is wrong.

The answer that scores

x \ln 0.5 > \ln 3

with

\ln 0.5 = - 0.693

, so

x < \frac{\ln 3}{\ln 0.5} = - 1.585

A student wroteA graph of

\ln y

against

x

has intercept 1.0986, so the constant

k

is 1.0986.

Why it loses marksThe linear form is

\ln y = \ln k + x \ln a

, so the intercept on the

\ln y

axis is

\ln k

. The value of

k

is the exponential of that intercept.

The answer that scores

\ln k = 1.0986

, so

k = e^{1.0986} = 3

A student wroteUsing

e^{\ln x} = x

for

x = - 5

to conclude that

e^{\ln (- 5)} = - 5

Why it loses marks

\ln (- 5)

is not defined: the argument of a logarithm must be positive. The inverse relationship only holds for

x > 0

The answer that scoresThere is no value of

e^{\ln (- 5)}

, because

\ln (- 5)

does not exist.

e^{\ln x} = x

holds only for

x > 0

Logarithmic and Exponential Functions

Syllabus coverage

A logarithm is an index written the other way round

The laws of logarithms

The exponential function e^x and the natural logarithm ln x

The graphs of y = e^x, y = ln x and y = e^kx

Solving equations in which the unknown is in the index

Reducing a relationship to linear form

How a Paper 3 logarithms question is built

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here