2 Exam Paper Ib SL Math Exponentials and Logs Logarithms

07/12/2018 2 Exam Paper - IB SL MAth - Exponentials and logs - Logarithms
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2 Exam Paper IB SL MAth Exponentials and logs Logarithms
Syllabus: IB(SL) Questions: 9 Time: 58 minutes Total Marks: 48
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2 Exam Paper IB SL MAth Exponentials and logs Logarithms
Questions: 9 Time: 58 minutes Total Marks: 48
Q1 ID: 638 [3 marks, 4 minutes]
Given
that e−x = 6, find the value of x, correct to 6 decimal places.
Given that ln(y + 6) − ln y = 0 6, find the value of y,

correct to 3 decimal places.
The function f is defined by

f :x = ln(15 − 5x) x 3 x
(a) Show that the inverse function of f is defined by
1
f −1 : x = 3 − ex
5
and write down the domain of f −1 .
−1
(b) Write down the range of f .
Find the exact coordinates of the points where the graph

of y = 7 ln(x + 1) − 8 crosses the coordinate axes.
Solve the equation ln(x + 11) = 7 + ln x,

giving your answer correct to 3 decimal places.
(a) Given that 3ex = 5, find the exact value of x.

(b) By substituting y = e show that the equation 3e + 35e−x = 26
x x

2
can be written as 3y − 26y + 35 = 0.
x −x
(c) Hence solve the equation 3e + 35e = 26 giving your answers as exact fractions.
2x2 − −7x − 4
(a) Simplify fully 2
x − −2x − 8
Given that ln(2x2 − −7x − 4) = 2 + ln(x2 − −2x − 8) x = −4
(b) find x in terms of e
Find the exact solutions to the equations

(a) ln(4x − 4) = 6
x
(b) 4 e 4x+6 = 9.
Find the exact solutions to the equations

(a)
ln x + ln 5 = ln 10,
x −x
(b) e + 28e = 11.
2 Exam Paper IB SL MAth Exponentials and logs Logarithms Mark Scheme
A1 ID: 638 [3 marks, 4 minutes]
e−x = 6 ex = 16
1
x = ln( 6 ) M 1A1

x = − 1 791759469228055
x = − 1 791759 A1
y+6
ln(y + 6) − ln y = 0 6 ln y
= 06 M 1A1
y+6 6
y = e0 6 1+ y = e0 6
6y = e0 6 − 1 M 1A1
y
6 = e0 16 −1
y = e0 66 −1 = 7 298 M 1A1
(a) y = ln(15 − 5x) ey = (15 − 5x) A1

5x = 15 − ey
x = 3 − 15 ey

f −1 (x) = 3 − 15 ex M 1A1
domain : x A1
(b) −1
range : f (x) 3 x A1
x = 0 y = 7 ln(1) − 8
y = −8 B1
y = 0 7 ln(x + 1) − 8 = 0 M1
7 ln(x + 1) = 8
ln(x + 1) = 87
8
x + 1 = e 7 A1
8
x = e 7 − 1 A1
ln(x + 11) = 7 + ln x ln(x + 11) − ln x = 7

x + 11
ln =7 M1
x
x + 11
= e7
x
x + 11 = e7 x A1
11 = (e7 − 1)x
11
x = 7 = 0 01 A1
e −1
5
(a) 3ex = 5 ex =
3
5
x = ln M 1A1
3
1
(b) y = ex 3y + 35 = 26
y
2
3y − 26y + 35 = 0 B1
2
(c) 3y − 26y + 35 = 0 (3y − 5)(y − 7) = 0
5
y = 7 B1
3
5
x = ln ln 7 M 1A1
3
2x2 − −7x − 4 (2x − 1)(x + 4)

(a) 2
= M 1B1
x − −2x − 8 (x − 2)(x + 4)
2x − 1
= A1
x−2
(b) equation ln(2x − −7x − 4) − ln(x2 − −2x − 8) = 2
2x − −7x − 4
ln 2 = 2 M1
x − −2x − 8
2x − 1
= e2 2x − 1 = (x − 2)e2 M1
x−2
2x − 1 = xe2 − 2e2
2 2
(2 − e )x = 1 − 2e
2
1 − 2e
x = M 1A1
2 − e2
(a) ln(4x − 4) = 6 4x − 4 = e6 B1

4x = e6 + 4
x = 14 (e6 + 4) M 1A1
x 4x+6
(b) 4e = 9 ln(4x e4x+6 ) = ln 9
ln 4x + ln e4x+6 = ln 9 M 1A1
x ln 4 + 4x + 6 = ln 9 B1
x(ln 4 + 4) = ln 9 − 6
ln 9 − 6
x = M 1A1
ln 4 + 4
(a) ln x + ln 5 = ln 10 ln 5x = ln 10

5x = 10
x = 2 M 1A1
(b) ex + 28e−x = 11 2x
e + 28 = 11e x

e2x − 11ex + 28 = 0 M 1A1
(ex − 7)(ex − 4) = 0
ex = 7 4
x = ln 7 ln 4 M 1A1

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07/12/2018 2 Exam Paper - IB SL MAth - Exponentials and logs - Logarithms

Given that ln(y + 6) − ln y = 0 6, ﬁnd the value of y,

The function f is deﬁned by

Find the exact coordinates of the points where the graph

Solve the equation ln(x + 11) = 7 + ln x,

(a) Given that 3ex = 5, ﬁnd the exact value of x.

Find the exact solutions to the equations

Find the exact solutions to the equations

(a) y = ln(15 − 5x) ey = (15 − 5x) A1

ln(x + 11) = 7 + ln x ln(x + 11) − ln x = 7

2x2 − −7x − 4 (2x − 1)(x + 4)

(a) ln(4x − 4) = 6 4x − 4 = e6 B1

(a) ln x + ln 5 = ln 10 ln 5x = ln 10

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