Sec 3 A-Math Exponential and logarithmic functions worksheet with answers

A free Sec 3 A-Math worksheet on exponential and logarithmic functions: 8 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 A-Math worksheet: Exponential and logarithmic functions

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

Four sketches A to D of exponential or logarithmic curves, each with its asymptote dashed and labelled.

The diagram shows four sketches,

A

B

C

and

D

, with their asymptotes. Which one is the graph of

y = 3e^{x} - 2

Answer[2]

Solve

\log_{3} (x - 4) = 2

Answer[2]

(a)Without using a calculator, find the value of

\log_{3} 81

Answer (a)[2]

(b)Without using a calculator, find the value of

\lg \dfrac{1}{10}

Answer (b)[2]

(a)Solve

3^{x} = 6

, giving your answer correct to 3 significant figures.

Answer (a)[2]

(b)Solve

3^{2x - 1} = 6

, giving your answer correct to 3 significant figures.

Answer (b)[2]

(a)Solve

5^{x - 2} = 125

Answer (a)[2]

(b)Solve

3^{2x} - 0(3^x) - 9 = 0

Answer (b)[3]

Section B

The diagram shows four sketches,

A

B

C

and

D

, with their asymptotes. It is given that

{f(x) = 2 + e^{x}}

and

{g(x) = \ln(x - 2)}

for

x > 2

Four sketches A to D: two exponential curves with horizontal asymptotes and two logarithmic curves with vertical asymptotes

(a)State which sketch shows the graph of

y = f(x)

Answer (a)[1]

(b)State which sketch shows the graph of

y = g(x)

Answer (b)[1]

(c)State the equation of the asymptote of the graph of

y = f(x)

Answer (c)[1]

(d)Find the coordinates of the point where the graph of

y = g(x)

meets the

x

-axis.

Answer (d)[1]

(e)By sketching the graphs of

y = f(x)

and

y = g(x)

on the same axes, state the number of solutions of the equation

f(x) = g(x)

Answer (e)[2]

The diagram shows part of the graph of

y = a + \log_{b} x

, where

a

and

b

are constants and

b > 1

. The graph passes through the points

(9, 3)

and

(27, 4)

The graph of y = a + log base b of x, rising slowly, passing through (9, 3) and (27, 4)

(a)Find the value of

b

Answer (a)[3]

(b)Find the value of

a

Answer (b)[1]

(c)Find the value of

x

when

y = 1

Answer (c)[2]

(d)Express

x

in terms of

y

Answer (d)[2]

A loaf of bread is taken out of an oven at

1735

. Its temperature,

T\,^\circ\text{C}

t

minutes later is modelled by

{\ln(T - 27) = \ln 133 - kt}

, where

k

is a positive constant.

(a)Show that

{T = 27 + 133e^{-kt}}

[2]

(b)After

10

minutes the temperature of the bread is

103\,^\circ\text{C}

. Find the value of

k

, giving your answer correct to 3 significant figures.

Answer (b)[2]

(c)State the temperature that the bread approaches after a long time.

Answer (c)°C[1]

(d)The bread can be sliced once its temperature has fallen to

62\,^\circ\text{C}

. Find the value of

t

at which this happens, giving your answer correct to 1 decimal place.

Answer (d)minutes[3]

(e)The bread is to be sliced at

1802

. Using your answer to part (d), state whether it will have cooled to

62\,^\circ\text{C}

by then.

Answer (e)[1]

END OF PAPER

Answers and worked solutions

Question 1

Answer: D[2]

x \to -\infty

e^{x} \to 0

, so

y \to -2

y = -2

y

increases as

x

increases (

3e^{x}

goes up).Sketch

D

Question 2

Answer:

(13) - 4 = 9 > 0

✓[2]

x - 4 = 3^{2}

x - 4 = 9

x = 13

(13) - 4 = 9 > 0

✓

Question 3

(a)Answer:

= 4

[2]

81 = 3^{4}

\log_{3} 81 = \log_{3}(3^{4})

= 4

(b)Answer:

= -1

[2]

\dfrac{1}{10} = 10^{-1}

\lg \dfrac{1}{10} = \lg(10^{-1})

= -1

Question 4

(a)Answer:

x = 1.63

[2]

\lg(3^{x}) = \lg 6

x \lg 3 = \lg 6

x = \dfrac{\lg 6}{\lg 3}

= 1.630930\ldots

x = 1.63

(b)Answer:

x = 1.32

[2]

\lg(3^{2x - 1}) = \lg 6

(2x - 1) \lg 3 = \lg 6

2x - 1 = \dfrac{\lg 6}{\lg 3}

= 1.630930\ldots

x = 1.32

Question 5

(a)Answer:

x = 5

[2]

125 = 5^{3}

5^{x - 2} = 5^{3}

x - 2 = 3

x = 5

(b)Answer:

x = 1

[3]

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

(3^{x})^2 - 9 = 0

y^2 - 9 = 0

(y + 3)(y - 3) = 0

y = -3

y = 3

3^x = -3

has no solution, since

3^x > 0

3^x = 3

3^x = 3^{1}

x = 1

Question 6

(a)Answer: Sketch

A

[1]

y = 2 + e^{x}

has asymptote

y = 2

and increases as

x

increases.This is sketch

A

(b)Answer: Sketch

C

[1]

y = \ln(x - 2)

is defined for

x > 2

: its asymptote is

x = 2

.This is sketch

C

(c)Answer:

y = 2

[1]

x \to -\infty

e^{x} \to 0

, so

y \to 2

.Asymptote:

y = 2

(d)Answer:

(3, 0)

[1]

\ln(x - 2) = 0

x - 2 = 1

x = 3

(e)Answer:

0

[2]

y = 2 + e^{x}

lies above its asymptote

y = 2

;

y = \ln(x - 2)

starts at its asymptote

x = 2

and increases slowly.The graphs meet nowhere.Number of solutions

= 0

Question 7

(a)Answer:

b = 3

[3]

a + \log_{b} 9 = 3

... (1)

a + \log_{b} 27 = 4

... (2)(2) − (1):

\log_{b} \dfrac{27}{9} = 1

b^{1} = 3

b = 3

(b)Answer:

a = 1

[1]

a = 3 - \log_{3} 9

= 3 - 2

= 1

(c)Answer:

x = 1

[2]

1 + \log_{3} x = 1

\log_{3} x = 0

x = 3^{0}

= 1

(d)Answer:

x = 3^{y - 1}

[2]

\log_{3} x = y - 1

x = 3^{y - 1}

Question 8

(a)[2]

\ln(T - 27) - \ln 133 = -kt

\ln \dfrac{T - 27}{133} = -kt

\dfrac{T - 27}{133} = e^{-kt}

T = 27 + 133e^{-kt}

(shown)

(b)Answer:

k = 0.0560

[2]

103 = 27 + 133e^{-10k}

e^{-10k} = \dfrac{4}{7}

10k = \ln \dfrac{7}{4}

k = 0.0560

(3 s.f.)

(c)Answer:

27\,^\circ\text{C}

°C[1]

t

becomes large,

e^{-kt}

approaches

0

.So

T

approaches

27\,^\circ\text{C}

(d)Answer:

t = 23.9

minutes[3]

62 = 27 + 133e^{-kt}

e^{-kt} = \dfrac{5}{19}

t = \dfrac{1}{k}\ln \dfrac{19}{5}

= 23.9

(e)Answer: Yes[1]

1802

27

minutes after

1735

27 > 23.9

, so the bread will have cooled to

62\,^\circ\text{C}

Practice questions by skill

Sec 3 A-Math Exponential and logarithmic functions worksheet with answers

In the print window, choose Save as PDF to download it.

Section B

The diagram shows four sketches,

A

B

C

and

D

, with their asymptotes. It is given that

{f(x) = 2 + e^{x}}

and

{g(x) = \ln(x - 2)}

for

x > 2

(a)State which sketch shows the graph of

y = f(x)

Answer (a)[1]

(b)State which sketch shows the graph of

y = g(x)

Answer (b)[1]

(c)State the equation of the asymptote of the graph of

y = f(x)

Answer (c)[1]

(d)Find the coordinates of the point where the graph of

y = g(x)

meets the

x

-axis.

Answer (d)[1]

(e)By sketching the graphs of

y = f(x)

and

y = g(x)

on the same axes, state the number of solutions of the equation

f(x) = g(x)

Answer (e)[2]

Answers and worked solutions

Question 1

Answer: D[2]

x \to -\infty

e^{x} \to 0

, so

y \to -2

y = -2

y

increases as

x

increases (

3e^{x}

goes up).Sketch

D

Question 2

Answer:

(13) - 4 = 9 > 0

✓[2]

x - 4 = 3^{2}

x - 4 = 9

x = 13

(13) - 4 = 9 > 0

✓

Question 3

(a)Answer:

= 4

[2]

81 = 3^{4}

\log_{3} 81 = \log_{3}(3^{4})

= 4

(b)Answer:

= -1

[2]

\dfrac{1}{10} = 10^{-1}

\lg \dfrac{1}{10} = \lg(10^{-1})

= -1

Question 4

(a)Answer:

x = 1.63

[2]

\lg(3^{x}) = \lg 6

x \lg 3 = \lg 6

x = \dfrac{\lg 6}{\lg 3}

= 1.630930\ldots

x = 1.63

(b)Answer:

x = 1.32

[2]

\lg(3^{2x - 1}) = \lg 6

(2x - 1) \lg 3 = \lg 6

2x - 1 = \dfrac{\lg 6}{\lg 3}

= 1.630930\ldots

x = 1.32

Question 5

(a)Answer:

x = 5

[2]

125 = 5^{3}

5^{x - 2} = 5^{3}

x - 2 = 3

x = 5

(b)Answer:

x = 1

[3]

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

(3^{x})^2 - 9 = 0

y^2 - 9 = 0

(y + 3)(y - 3) = 0

y = -3

y = 3

3^x = -3

has no solution, since

3^x > 0

3^x = 3

3^x = 3^{1}

x = 1

Question 6

(a)Answer: Sketch

A

[1]

y = 2 + e^{x}

has asymptote

y = 2

and increases as

x

increases.This is sketch

A

(b)Answer: Sketch

C

[1]

y = \ln(x - 2)

is defined for

x > 2

: its asymptote is

x = 2

.This is sketch

C

(c)Answer:

y = 2

[1]

x \to -\infty

e^{x} \to 0

, so

y \to 2

.Asymptote:

y = 2

(d)Answer:

(3, 0)

[1]

\ln(x - 2) = 0

x - 2 = 1

x = 3

(e)Answer:

0

[2]

y = 2 + e^{x}

lies above its asymptote

y = 2

;

y = \ln(x - 2)

starts at its asymptote

x = 2

and increases slowly.The graphs meet nowhere.Number of solutions

= 0

Question 7

(a)Answer:

b = 3

[3]

a + \log_{b} 9 = 3

... (1)

a + \log_{b} 27 = 4

... (2)(2) − (1):

\log_{b} \dfrac{27}{9} = 1

b^{1} = 3

b = 3

(b)Answer:

a = 1

[1]

a = 3 - \log_{3} 9

= 3 - 2

= 1

(c)Answer:

x = 1

[2]

1 + \log_{3} x = 1

\log_{3} x = 0

x = 3^{0}

= 1

(d)Answer:

x = 3^{y - 1}

[2]

\log_{3} x = y - 1

x = 3^{y - 1}

Question 8

(a)[2]

\ln(T - 27) - \ln 133 = -kt

\ln \dfrac{T - 27}{133} = -kt

\dfrac{T - 27}{133} = e^{-kt}

T = 27 + 133e^{-kt}

(shown)

(b)Answer:

k = 0.0560

[2]

103 = 27 + 133e^{-10k}

e^{-10k} = \dfrac{4}{7}

10k = \ln \dfrac{7}{4}

k = 0.0560

(3 s.f.)

(c)Answer:

27\,^\circ\text{C}

°C[1]

t

becomes large,

e^{-kt}

approaches

0

.So

T

approaches

27\,^\circ\text{C}

(d)Answer:

t = 23.9

minutes[3]

62 = 27 + 133e^{-kt}

e^{-kt} = \dfrac{5}{19}

t = \dfrac{1}{k}\ln \dfrac{19}{5}

= 23.9

(e)Answer: Yes[1]

1802

27

minutes after

1735

27 > 23.9

, so the bread will have cooled to

62\,^\circ\text{C}