What is 6.3: Logarithms and ln x?

For a>0 with a1, _a x=y means exactly that a^y=x; a real logarithm requires x>0.

How many marks are 6.3 questions worth?

6.3 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 6.3 appeared in?

We have not yet indexed a verified past-paper appearance for 6.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

6.3

Know and use the definition of logₐx as the inverse of aˣ, where a is positive and x ≥ 0; know and use the function ln x and its graph; know and use ln x as the inverse function of eˣ.

Draft — not yet indexed

Logarithms and ln x

Worked answers and methods for 6.3 on Edexcel A-level Maths 9MA0.

Explanation

For $a>0$ with $a\neq1$ , $\log_a x=y$ means exactly that $a^y=x$ ; a real logarithm requires $x>0$ .
The graph of $y=\log_a x$ is the reflection of $y=a^x$ in $y=x$ , so its domain is $x>0$ , range is $\mathbb{R}$ and vertical asymptote is $x=0$ .
Natural logarithms use base $e$ , giving the inverse relations $\ln(e^x)=x$ for every real $x$ and $e^{\ln x}=x$ for $x>0$ .
The specification's inverse relationship does not make $\ln0$ defined: $e^x$ approaches zero but never equals it, so logarithm arguments must be strictly positive.

Worked example

The function $f(x)=e^{x-3}+2$ . Find $f^{-1}(x)$ and state the domain of the inverse.

1.Write $y=e^{x-3}+2$ , so $y-2=e^{x-3}$ .
2.Taking natural logarithms gives $\ln(y-2)=x-3$ , hence $x=\ln(y-2)+3$ .
3.Interchanging labels gives $f^{-1}(x)=\ln(x-2)+3$ .
4.The range of $f$ is $y>2$ , so the inverse domain is $x>2$ .

Answer: $f^{-1}(x)=\ln(x-2)+3$ Domain $x>2$

Common mistakes

Don't include zero in the domain of $\ln x$ , even though $e^x$ approaches but never reaches zero.
Don't find an inverse formula and fail to interchange the exponential function's range with the logarithm's domain.

Exam tip

Before taking a logarithm, require its argument to be strictly positive and state the resulting domain.

Worked practice

Tier 1 · Easy

Solve

\ln x=2

exactly.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$x=e^2$	2
Notes
Apply $e$ to both sides: $e^{\ln x}=e^2$ , hence $x=e^2$ .

(2 marks)

Tier 2 · Standard

For

y=\ln(x-2)+1

, state the domain, range and vertical asymptote, and find the exact

x

-intercept.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Domain $x>2$ Range $y\in\mathbb{R}$ Vertical asymptote $x=2$ $x$ -intercept $\left(2+e^{-1},0\right)$	4
Notes
The logarithm requires $x-2>0$ , so the domain is $x>2$ and the vertical asymptote is $x=2$ . A translation does not restrict the range of the logarithm, so the range is all real numbers. At the $x$ -intercept, $\ln(x-2)+1=0$ , so $x-2=e^{-1}$ and $x=2+e^{-1}$ .

(4 marks)

Tier 3 · Hard

For

g(x)=\ln(4-x^2)

, determine the domain and range, and identify the maximum point.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Domain $-2<x<2$ Range $-\infty<g(x)\leq\ln4$ Maximum point $(0,\ln4)$	5
Notes
The logarithm requires $4-x^2>0$ , so $x^2<4$ and $-2<x<2$ . The argument $4-x^2$ has maximum $4$ at $x=0$ , giving the maximum value $\ln4$ . As $x\to2^-$ or $x\to-2^+$ , the argument tends to $0^+$ and $g(x)\to-\infty$ . Therefore the range is $(-\infty,\ln4]$ .

(5 marks)

Tier 1 · Easy

Find the exact value of

\log_5\!\left(\dfrac1{125}\right)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$-3$	2
Notes
Since $125=5^3$ , $1/125=5^{-3}$ . By the definition of a logarithm, $\log_5(5^{-3})=-3$ .

(2 marks)

Tier 2 · Standard

Let

f(x)=3-e^{-x}

for

x\in\mathbb{R}

. Find

f^{-1}(x)

and state the domain of

f^{-1}

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$f^{-1}(x)=-\ln(3-x)$ Domain $x<3$	4
Notes
Write $y=3-e^{-x}$ . Then $e^{-x}=3-y$ , so $-x=\ln(3-y)$ and $x=-\ln(3-y)$ . Interchanging $x$ and $y$ gives $f^{-1}(x)=-\ln(3-x)$ . Since $e^{-x}>0$ , the range of $f$ is $y<3$ , which is the inverse domain.

(4 marks)

Tier 3 · Hard

The function

f(x)=\ln\!\left(\dfrac{x+2}{4-x}\right)

is defined on its largest possible interval. Find its domain, range and vertical asymptotes. Hence find

f^{-1}(x)

and state its domain.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Domain $-2<x<4$ Range $\mathbb{R}$ Vertical asymptotes $x=-2$ and $x=4$ $f^{-1}(x)=\dfrac{4e^x-2}{e^x+1}$ , with domain $x\in\mathbb{R}$	6
Notes
The quotient is positive on $-2<x<4$ , which is the required interval. It tends to $0^+$ as $x\to-2^+$ and to $+\infty$ as $x\to4^-$ , so $f$ has the two vertical asymptotes and range $\mathbb{R}$ . If $y=\ln((x+2)/(4-x))$ , then $e^y(4-x)=x+2$ . Hence $x=(4e^y-2)/(e^y+1)$ , giving the stated inverse after interchanging variables. Its domain is the range of $f$ , namely $\mathbb{R}$ .

(6 marks)

Tier 2 · Standard

Given that

\log_a8=\dfrac32

, find

a

. Hence solve

\log_a(x-1)=-\dfrac12

, giving the exact value of

x

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$a=4$ $x=\dfrac32$	4
Notes
The definition of a logarithm gives $a^{3/2}=8$ , so $(\sqrt a)^3=8$ . Since a logarithm base is positive, $\sqrt a=2$ and $a=4$ . Then $\log_4(x-1)=-1/2$ gives $x-1=4^{-1/2}=1/2$ , so $x=3/2$ , which satisfies $x-1>0$ .

(4 marks)

Tier 3 · Hard

Sketch the graph of

y=|\ln x|

for

x>0

, marking its

x

-intercept, vertical asymptote and minimum point. State its range. Hence solve

|\ln x|=\ln7

exactly.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$y=-\ln x$ for $0<x\leq1$ and $y=\ln x$ for $x\geq1$ , with $x$ -intercept and minimum point $(1,0)$ and vertical asymptote $x=0$ Range $y\geq0$ $x=\dfrac17$ or $x=7$	6
Notes
For $0<x<1$ , $\ln x<0$ , so $\|\ln x\|=-\ln x$ ; for $x\geq1$ , $\|\ln x\|=\ln x$ . The two branches meet at $(1,0)$ , the minimum point and only $x$ -intercept. As $x\to0^+$ , $\|\ln x\|\to\infty$ , so $x=0$ is a vertical asymptote; as $x\to\infty$ , $\|\ln x\|\to\infty$ . Hence the range is $y\geq0$ . Finally, $\|\ln x\|=\ln7$ gives $\ln x=\pm\ln7$ , so $x=7$ or $x=1/7$ .

(6 marks)

Tier 3 · Hard

A function

f

has inverse

f^{-1}(x)=3^{2x-1}+4

for

x\in\mathbb{R}

. Find

f(x)

. State the domain, range and vertical asymptote of

f

, and find its exact

x

-intercept.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$f(x)=\dfrac{1+\log_3(x-4)}2$ Domain $x>4$ , range $\mathbb{R}$ and vertical asymptote $x=4$ $x$ -intercept $\left(\dfrac{13}{3},0\right)$	6
Notes
Write $y=3^{2x-1}+4$ . Then $y-4=3^{2x-1}$ , so $\log_3(y-4)=2x-1$ and $x=(1+\log_3(y-4))/2$ . Interchanging variables gives the stated $f$ . Its logarithm requires $x>4$ , its range is all real numbers, and its vertical asymptote is $x=4$ . For the $x$ -intercept, $1+\log_3(x-4)=0$ , so $x-4=1/3$ and $x=13/3$ .

(6 marks)

Q10

Tier 3 · Hard

10.

A point

P

lies on the curve

y=3^x

, and

Q

is the reflection of

P

in the line

y=x

. Prove that

Q

lies on

y=\log_3x

. Given that the midpoint of

PQ

M\left(\dfrac{11}{2},\dfrac{11}{2}\right)

, find the coordinates of

P

and

Q

without using numerical methods. Hence find the equation of

PQ

and its exact length.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$P=(2,9)$ and $Q=(9,2)$ $PQ$ : $y=11-x$ $\|PQ\|=7\sqrt2$	6
Notes
Write $P=(p,3^p)$ . Reflection in $y=x$ interchanges the coordinates, so $Q=(3^p,p)$ . Since $\log_3(3^p)=p$ , the point $Q$ lies on $y=\log_3x$ . The midpoint condition gives $p+3^p=11$ . The value $p=2$ satisfies this equation, and it is the unique solution because both $p$ and $3^p$ are strictly increasing functions of $p$ . Hence $P=(2,9)$ and $Q=(9,2)$ . The line through them has gradient $-1$ , so its equation is $y=11-x$ . Finally, $\|PQ\|=\sqrt{(9-2)^2+(2-9)^2}=7\sqrt2$ .

(6 marks)

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