Free worked solutions · 2026 Edition

Chapter 2: Exponentials and logarithms worked solutions

Worked solutions to the practice questions in Chapter 2 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[6 marks]

In this question x\ne 0.

(a) The constants a and p are such that \left(ax^{p}\right)^{3}=64x^{-6} for all x\ne 0. Find the value of a and the value of p.

[3]

(b) The integer k is such that \dfrac{2x^{k}\times 5x^{-3}}{x^{4}}=10x^{-2} for all x\ne 0. Find k.

[2]

(c) The integer m is such that \left(x^{m}\right)^{-2}\times x^{8}=1 for all x\ne 0. Find m.

[1]

Show solutionHide solution

(a) a^{3}x^{3p}=64x^{-6}

a^{3}=64 \Rightarrow a=4

3p=-6 \Rightarrow p=-2

(b) 10x^{k-3-4}=10x^{-2}, so k-7=-2

k=5

(c) x^{8-2m}=x^{0}, so 8-2m=0, m=4

Final answer

(a) a=4, p=-2
(b) k=5
(c) m=4

Question 2

[8 marks]

Find the exact value of each of the following, without using a calculator.

(a) \log_{4}2+\log_{4}32

[2]

(b) \log_{2}48-\log_{2}3

[2]

(c) 2\log 5+\log 4

[2]

(d) \log_{5}\sqrt[3]{25}

[2]

Show solutionHide solution

(a) \log_{4}(2\times 32)=\log_{4}64

(b) \log_{2}\dfrac{48}{3}=\log_{2}16

(c) \log 25+\log 4=\log 100

(d) \log_{5}\left(5^{\frac{2}{3}}\right)

=\dfrac{2}{3}

Final answer

(a) 3
(b) 4
(c) 2
(d) \frac23

Question 3

[7 marks]

Solve the following equations, giving each answer exactly.

(a) 6^{x}=17. Give your answer in the form \dfrac{\ln m}{\ln n}.

[1]

(b) 2^{x+1}=13. Give your answer in the form \log_{2}k.

[2]

(c) e^{3x-2}=5

[2]

(d) 10^{1-x}=0.4. Give your answer in the form \log k, where k\in\mathbb{Z}.

[2]

Show solutionHide solution

(a) x=\log_{6}17=\dfrac{\ln 17}{\ln 6}

(b) x+1=\log_{2}13

x=\log_{2}13-\log_{2}2=\log_{2}\dfrac{13}{2}

(c) 3x-2=\ln 5

x=\dfrac{2+\ln 5}{3}

(d) 1-x=\log 0.4

x=\log 10-\log 0.4=\log 25

Final answer

(a) \frac{\ln17}{\ln6}
(b) \log_2\frac{13}{2}
(c) \frac{2+\ln5}{3}
(d) \log25

Question 4

[8 marks]

Let \ln 3=p and \ln 5=q.

Express each of the following in terms of p and q.

(a) \ln 45

[2]

(b) \ln 0.6

[2]

(c) \ln\left(\dfrac{\sqrt{5}}{9e}\right)

[2]

(d) \log_{5}75

[2]

Show solutionHide solution

(a) \ln\left(3^{2}\times 5\right)=2\ln 3+\ln 5

=2p+q

(b) \ln\dfrac{3}{5}

=p-q

(c) \dfrac{1}{2}\ln 5-2\ln 3-\ln e

=\dfrac{1}{2}q-2p-1

(d) \log_{5}75=\dfrac{\ln 75}{\ln 5}=\dfrac{\ln 3+2\ln 5}{\ln 5}

=\dfrac{p+2q}{q}

Note: accept \dfrac{p}{q}+2.

Final answer

(a) 2p+q
(b) p-q
(c) \frac q2-2p-1
(d) \frac{p+2q}{q}

Question 5

[9 marks]

Solve the following equations.

(a) \log_{3}x-\log_{9}x=\dfrac{3}{2}

[4]

(b) \log_{2}x=\log_{4}(3x+4)

[5]

Show solutionHide solution

(a) \log_{9}x=\dfrac{\log_{3}x}{\log_{3}9}=\dfrac{1}{2}\log_{3}x M1A1

\dfrac{1}{2}\log_{3}x=\dfrac{3}{2}\Rightarrow\log_{3}x=3

x=27

(b) \log_{4}(3x+4)=\dfrac{1}{2}\log_{2}(3x+4)

2\log_{2}x=\log_{2}(3x+4)\Rightarrow x^{2}=3x+4

(x-4)(x+1)=0

x=4

x=-1 is rejected because \log_{2}x needs x>0

Final answer

(a) x=27
(b) x=4 (x=-1 rejected)

Question 6

[7 marks]

Solve the following equations, giving exact answers.

(a) e^{2x}-7e^{x}+10=0

[3]

(b) e^{2x}+e^{x}-6=0

[4]

Show solutionHide solution

(a) \left(e^{x}-2\right)\left(e^{x}-5\right)=0

M1A1

x=\ln 2 or x=\ln 5

Note: M1 for treating the equation as a quadratic in e^x (e.g. y^2-7y+10=0), A1 for the correct factors or roots e^x=2,\ 5.

(b) \left(e^{x}+3\right)\left(e^{x}-2\right)=0

e^{x}=-3 or e^{x}=2

e^{x}=-3 is impossible because e^{x}>0 for all x

x=\ln 2

Final answer

(a) x=\ln2 or x=\ln5
(b) x=\ln2 (e^x=-3 rejected)

Question 7

[7 marks]

Let f(x)=7^x and h(x)=7^{-x}, both for x\in\mathbb{R}.

(a) Write down the coordinates of the point where each graph meets the y-axis.

[1]

(b) Write down the equation of the horizontal asymptote shared by both graphs.

[1]

(c) Write down the domain and the range of f.

[2]

(d) State which of the two functions is decreasing.

[1]

(e) Explain why the equation 7^x=-2 has no solution.

[1]

(f) Find the value of x for which h(x)=49.

[1]

Show solutionHide solution

(a) both graphs pass through (0,\ 1)

(b) y=0

(c) domain x\in\mathbb{R}

range f(x)>0

(d) h (since h(x)=\left(\tfrac{1}{7}\right)^x with 0<\tfrac{1}{7}<1)

(e) the range of f is f(x)>0, so 7^x is never negative

(f) 7^{-x}=7^{2} \Rightarrow x=-2

Final answer

(a) (0,1) for both
(b) y=0
(c) domain \mathbb{R}, range f(x)>0
(d) h
(e) 7^x>0 for every x
(f) x=-2

Question 8

[9 marks]

Let g(x)=\ln(2x-4)+1, for x>2.

(a) Find g\left(\tfrac{5}{2}\right).

[1]

(b) Find g^{-1}(x).

[3]

(c) Write down the range of g^{-1}.

[1]

(d) Write down the equation of the horizontal asymptote of the graph of g^{-1}.

[1]

(e) On the same axes, sketch the graphs of y=g(x) and y=g^{-1}(x), showing their asymptotes and the line y=x.

[3]

Show solutionHide solution

(a) \ln 1+1=1

(b) x=\ln(2y-4)+1 \Rightarrow 2y-4=e^{x-1}

M1A1

g^{-1}(x)=\dfrac{e^{x-1}+4}{2}

Note: award M1 for interchanging x and y (at any stage) and an attempt to undo the logarithm, A1 for the correct exponential equation.

(c) g^{-1}(x)>2 (the domain of g)

(d) y=2

(e)

(Sketch of both curves on one pair of axes, with the asymptotes x=2 and y=2 and the line y=x.)

g: increasing, vertical asymptote x=2, through \left(\tfrac{5}{2},1\right)

g^{-1}: increasing, horizontal asymptote y=2, through \left(1,\tfrac{5}{2}\right)

the two curves are reflections of each other in y=x (and do not meet it)

Final answer

(a) 1
(b) g^{-1}(x)=\frac{e^{x-1}+4}{2}
(c) g^{-1}(x)>2
(d) y=2
(e) sketch: see the online solutions

Question 9

[7 marks]

The number of cells in a laboratory culture is modelled by N(t)=500\times 2^{t/3}, where t is the time in hours after the culture was started.

(a) Write down the number of cells when the culture was started.

[1]

(b) Find the number of cells after 6 hours.

[2]

(c) Explain what the number 3 in the exponent tells you about the culture.

[1]

(d) Find the time at which the number of cells reaches 16\,000.

[3]

Show solutionHide solution

(a) 500

(b) 500\times 2^{2}

=2000

(c) the number of cells doubles every 3 hours

(d) 500\times 2^{t/3}=16\,000 \Rightarrow 2^{t/3}=32

\dfrac{t}{3}=5

t=15 hours

Final answer

(a) 500
(b) 2000
(c) the number of cells doubles every 3 hours
(d) t=15 hours

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 10

[7 marks]

Let \log_{5}2=a and \log_{5}3=b. Find expressions, in terms of a and b, for

(a) \log_{5}12;

[2]

(b) \log_{5}0.72;

[3]

(c) \log_{5}\sqrt{\dfrac{20}{9}}.

[2]

Show solutionHide solution

(a) 12=2^2\times3, so \log_512=2\log_52+\log_53

[1]

=2a+b.

[1]

(b) 0.72=\dfrac{72}{100}=\dfrac{2^3\times3^2}{100}, so \log_50.72=3a+2b-\log_5100.

[1]

\log_5100=\log_5(5^2\times2^2)=2+2a.

[1]

\log_50.72=3a+2b-2-2a=a+2b-2.

[1]

(c) \log_5\sqrt{\tfrac{20}{9}}=\tfrac12\left(\log_520-\log_59\right) with \log_520=2a+1 and \log_59=2b.

[1]

=\tfrac12(2a+1-2b)=a+\tfrac12-b.

[1]

Final answer

(a) 2a+b
(b) a+2b-2
(c) a+\frac12-b

Question 11

[6 marks]

Positive numbers x and y and a base c>1 satisfy

\log_{c}(xy)=5\qquad\text{and}\qquad\log_{c}\!\left(\frac{x}{y^{2}}\right)=2.

(a) Find \log_{c}x and \log_{c}y.

[4]

(b) Hence find \log_{c}\!\left(x^{2}\sqrt{y}\right).

[2]

Show solutionHide solution

(a) Let u=\log_cx and v=\log_cy.

\log_c(xy)=u+v=5.

[1]

\log_c\!\left(xy^{-2}\right)=u-2v=2.

[1]

Subtract: 3v=3, so v=1.

[1]

u=4.

[1]

(b) \log_c\!\left(x^2\sqrt y\right)=2u+\tfrac12v

[1]

=8+\tfrac12=\dfrac{17}{2}.

[1]

Final answer

(a) \log_cx=4, \log_cy=1
(b) \frac{17}{2}

Question 12

[6 marks]

A calibration reading z must obey \log_{25}(z+11)=\log_5(z-7). Find every valid real reading and justify exclusions.

[6]

Show solutionHide solution

The right-hand logarithm restricts the reading to z>7. Change of base makes the left side \tfrac12\log_5(z+11). Multiplication by two and the power law give z+11=(z-7)^2. This is z^2-15z+38=0, whose roots are (15\pm\sqrt{73})/2. The minus root lies below 7 and is excluded. The plus root exceeds 7 and satisfies the original positive arguments. Hence

z=\frac{15+\sqrt{73}}2.

[6]

Final answer

z=\frac{15+\sqrt{73}}{2}

Question 13

[7 marks]

Consider the equation e^{2x}-4e^{x}+m=0, where m\in\mathbb{R}.

(a) Find the set of values of m for which the equation has two distinct real solutions.

[4]

(b) Find the set of values of m for which the equation has exactly one real solution.

[3]

Show solutionHide solution

(a) Let u=e^x>0: u^2-4u+m=0. Each positive root u gives exactly one x.

[1]

Two distinct real roots: \Delta=16-4m>0\Rightarrow m<4.

[1]

Both roots must be positive: sum =4>0 and product =m>0.

[1]

0<m<4.

[1]

(b) m=4: repeated root u=2>0, one solution.

[1]

m\le0: product of roots m\le0, so one root is positive and the other is negative or zero (for m=0: u=0 or 4). Only the positive root is usable.

[1]

So exactly one solution when m=4 or m\le0.

[1]

Final answer

(a) 0<m<4
(b) m=4 or m\le0

Question 14

[8 marks]

Find all values of k\in\mathbb{R} for which the equation \ln(x+k)=2\ln x has exactly one real solution.

[8]

Show solutionHide solution

The equation requires x>0 and x+k>0. For x>0, 2\ln x=\ln x^2, so the equation is x+k=x^2.

[1]

That is x^2-x-k=0 with x>0 (then x+k=x^2>0 holds automatically).

[1]

Discriminant \Delta=1+4k.

[1]

k<-\tfrac14: \Delta<0, no solutions.

[1]

k=-\tfrac14: repeated root x=\tfrac12>0, exactly one solution.

[1]

k>0: the product of the roots is -k<0, so one root is positive and one negative: exactly one valid solution.

[1]

k=0: x^2-x=0, roots 0 and 1; x=0 is not allowed, so exactly one solution (x=1).

[1]

-\tfrac14<k<0: two distinct roots with sum 1>0 and product -k>0, so both positive and valid: two solutions, not allowed.

[1]

Exactly one solution if and only if k=-\tfrac14 or k\ge0.

Final answer

k=-\frac14 or k\ge0

Question 15

[7 marks]

A fabric test models colour strength by C(t)=C_0e^{kt}, with days t\ge0. After four days, strength has fallen by 27\%.

(a) Calculate k to three significant figures.

[2]

(b) Identify when strength is below 18\% of its initial level.

[3]

(c) For C_0=63, calculate strength after eleven days to three significant figures.

[2]

Show solutionHide solution

(a) Retained strength at day four is 0.73C_0. Therefore k=\ln(0.73)/4=-0.0787 per day.

[2]

(b) The inequality e^{kt}<0.18 becomes kt<\ln(0.18). Since k<0, division reverses the sign: t>4\ln(0.18)/\ln(0.73), approximately 21.8 days.

[3]

(c) Substitution gives C(11)=63(0.73)^{11/4}=26.5 strength units.

[2]

Final answer

(a) k=-0.0787
(b) t>21.8 days
(c) 26.5

Question 16

[8 marks]

The magnitude m of a star is defined by m=-2.5\log_{10}b, where b is the brightness of the star measured relative to a standard star of brightness 1. A larger magnitude means a fainter star.

(a) A star has brightness 0.0032. Find its magnitude, correct to three significant figures.

[2]

(b) A star has magnitude 1.4. Find its brightness, correct to three significant figures.

[2]

(c) Star A is 100 times as bright as star B. Show that the magnitude of star B is exactly 5 more than the magnitude of star A.

[2]

(d) Two stars differ in magnitude by 1. Find the ratio of the brightness of the brighter star to the brightness of the fainter star, correct to three significant figures.

[2]

Show solutionHide solution

(a) m=-2.5\log_{10}0.0032.

[1]

=6.24 (6.2371\ldots).

[1]

(b) -2.5\log_{10}b=1.4\Rightarrow\log_{10}b=-0.56.

[1]

b=10^{-0.56}=0.275 (0.27542\ldots).

[1]

(c) b_A=100\,b_B. Then m_B-m_A=-2.5\left(\log_{10}b_B-\log_{10}b_A\right)=-2.5\log_{10}\dfrac{b_B}{b_A}.

[1]

=-2.5\log_{10}\dfrac1{100}=-2.5\times(-2)=5. [1]

(d) Let the brighter star have brightness b_1 and magnitude m_1, and the fainter star b_2 and m_2. The fainter star has the larger magnitude, so m_2-m_1=1 and -2.5\log_{10}\dfrac{b_2}{b_1}=1.

[1]

\log_{10}\dfrac{b_1}{b_2}=0.4, so \dfrac{b_1}{b_2}=10^{0.4}=2.51 (2.5119\ldots).

[1]

Final answer

(a) 6.24
(b) 0.275
(c) shown
(d) 2.51

Question 17

[7 marks]

Imogen invests \$12\,000 in an account that pays 4.2\% per annum, compounded monthly.

(a) Find the value of the investment after 8 years, to the nearest dollar.

[2]

(b) Find the least number of complete months after which the value first exceeds \$18\,500.

[3]

A second account pays the same nominal rate of 4.2\% per annum, but compounded continuously, with the value after t years given by A=12\,000\,e^{0.042t}.

(c) Find the value of this investment after 8 years, to the nearest dollar.

[2]

Show solutionHide solution

(a) Monthly rate \dfrac{4.2}{12}=0.35\%, so A=12\,000(1.0035)^{96}.

[1]

A=\$16\,782 (16\,782.22\ldots).

[1]

(b) 12\,000(1.0035)^n>18\,500\Rightarrow(1.0035)^n>\dfrac{37}{24}.

[1]

n>\dfrac{\ln(37/24)}{\ln1.0035}=123.9\ldots

[1]

The least number of complete months is 124.

[1]

(c) A=12\,000e^{0.042\times8}=12\,000e^{0.336}.

[1]

A=\$16\,792 (16\,792.07\ldots).

[1]

Final answer

(a) \$16\,782
(b) 124 months
(c) \$16\,792

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 18

[5 marks]

Solve the equation

2\log_{27}(9x) - \log_{9} x = 1, \qquad x>0.

[5]

Show solutionHide solution

Convert every logarithm to base 3, using \log_{27}y=\dfrac{\log_3 y}{3} and \log_9 y=\dfrac{\log_3 y}{2}:

\frac{2}{3}\log_3(9x)-\frac12\log_3 x=1

Split \log_3(9x)=2+\log_3 x and write L=\log_3 x:

\frac{2}{3}(2+L)-\frac{1}{2}L=1

\frac43+\frac{2}{3}L-\frac12L=1

\frac{L}{6}=-\frac13 \implies L=-2

Hence \log_3 x=-2, so

x=3^{-2}=\frac19

x=\dfrac{1}{9}

[5]

Final answer

x=\frac19

Question 19

[5 marks]

Let \log_{3} 2 = a and \log_{3} 5 = b.

(a) Express \log_{3} 60 in terms of a and b.

[3]

(b) Express \log_{15} 8 in terms of a and b.

[2]

Show solutionHide solution

(a) Factorise 60 = 2^{2}\times 3\times 5 and use the product and power laws:

\log_{3}60 = 2\log_{3}2 + \log_{3}3 + \log_{3}5

= 2a + 1 + b

[3]

(b) Change to base 3, with 8=2^{3} and 15 = 3\times5:

\log_{15}8 = \frac{\log_{3}8}{\log_{3}15} = \frac{3\log_{3}2}{\log_{3}3+\log_{3}5}

= \dfrac{3a}{1 + b}

[2]

Final answer

(a) 2a+1+b
(b) \frac{3a}{1+b}

Question 20

[6 marks]

Find the set of values of k > 0 for which the equation

e^{2x} - 10\,e^{x} + \ln(3k) = 0

has exactly two distinct real solutions for x.

[6]

Show solutionHide solution

Let u = \mathrm{e}^{x}, so u>0. Each positive u gives exactly one x, hence the original equation has two solutions exactly when

u^{2} - 10u + \ln(3k) = 0

has two distinct positive roots.

Distinct roots: the discriminant must be positive,

100 - 4\ln(3k) > 0 \implies \ln(3k) < 25.

Both roots positive: the sum of roots is 10>0 already, so we need the product positive:

\ln(3k) > 0 \implies 3k > 1.

Combining the two conditions,

0 < \ln(3k) < 25 \implies 1 < 3k < \mathrm{e}^{25}

\dfrac{1}{3} < k < \dfrac{\mathrm{e}^{25}}{3}

[6]

Final answer

\frac13<k<\frac{e^{25}}{3}

Question 21

[5 marks]

Let h(x) = \ln\!\left(2x^{2} - 4mx + 3m + 5\right), where m is a real constant.

The largest possible domain of h is all of \mathbb{R}. Find the set of possible values of m.

[5]

Show solutionHide solution

The logarithm is defined only for positive arguments, so we need

2x^{2} - 4mx + 3m + 5 > 0 \quad\text{for every real } x.

The parabola opens upwards (coefficient 2>0), so it stays strictly above the x-axis exactly when it has no real roots, i.e. when its discriminant is negative:

\Delta = (-4m)^{2} - 4(2)(3m+5) = 16m^{2} - 24m - 40 < 0

Dividing by 8:

2m^{2} - 3m - 5 < 0

(2m - 5)(m + 1) < 0

The product is negative between the roots m=-1 and m=\frac52:

-1 < m < \dfrac{5}{2}

[5]

Final answer

-1<m<\frac52

Question 22

[14 marks]

The function f is given by f(x) = \log_{2}(8x), for x > 0.

(a) Evaluate (i) f(8); (ii) f\left(\dfrac{1}{8}\right).

[3]

(b) Obtain an expression for f^{-1}(x).

[4]

(c) Hence, or otherwise, evaluate f^{-1}(0).

[1]

The graph of y = f\left(64x^{3}\right) may be produced from the graph of y = \log_{2} x by a

stretch together with a translation.

(d) Describe these two transformations, stating the order in which they are applied.

[6]

Show solutionHide solution

(a)(i) f(8) = \log_{2}(8 \times 8) = \log_{2} 64 = \mathbf{6} M1A1

(ii) f\left(\tfrac18\right) = \log_{2}\left(8 \times \tfrac18\right) = \log_{2} 1 = \mathbf{0}

A1[3]

(b) Let y = \log_{2}(8x) and interchange the variables:

x = \log_{2}(8y)

Rewriting in exponential form:

2^{x} = 8y \implies y = \frac{2^{x}}{8}

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

A1[4]

(c) f^{-1}(0) = 2^{-3} = \mathbf{\dfrac{1}{8}}

A1[1]

(d) Simplify the composite first:

f\left(64x^{3}\right) = \log_{2}\left(8 \times 64x^{3}\right) = \log_{2}\left(512x^{3}\right)

Splitting with the addition and power laws:

= \log_{2} 512 + \log_{2} x^{3} = 9 + 3\log_{2} x

So, starting from y = \log_{2} x:

1. a vertical stretch with scale factor3;

2. followed by a vertical translation of9 units upwards.

(The order matters: translating first and then stretching would give 3\log_2 x + 27.)

[6]

Final answer

(a) 6; 0
(b) f^{-1}(x)=2^{x-3}
(c) \frac18
(d) vertical stretch, scale factor 3, then translation 9 up

Question 23

[6 marks]

The number of subscribers to an online magazine t years after 1 March 2016 is modelled by

S(t) = 8200e^{kt},

where k < 0 and t \geq 0.

Between 1 March 2016 and 1 March 2022 the number of subscribers fell by 18\%.

Use this model to estimate the number of subscribers on 1 March 2035, giving your answer to the nearest whole number.

[6]

Show solutionHide solution

After 6 years the number of subscribers is 82\% of 8200:

8200e^{6k} = 0.82\times 8200 \implies e^{6k} = 0.82 \implies k = \frac{\ln 0.82}{6} = -0.0331\ldots \quad \textbf{M1A1A1}

On 1 March 2035, t = 19:

S(19) = 8200e^{19k} = 8200\times 0.82^{\frac{19}{6}} \quad \textbf{M1A1}

S(19) = 4374 \text{ subscribers} \quad \textbf{A1}\quad\textbf{[6]}

Final answer

k=\frac{\ln0.82}{6}=-0.0331; 4374 subscribers

Question 24

[7 marks]

A tractor costs \$52\,000 new. Its value falls by 30\% in its first year and then by 6\% of its current value each year after that.

(a) Find its value five years after purchase, to the nearest dollar.

[3]

(b) Find the number of complete years from purchase after which its value is first below half of the new price.

[4]

Show solutionHide solution

(a) After year 1: 52\,000 \times 0.70 = 36\,400

Four further years: 36\,400 \times 0.94^{4}

= \$28419

A1[3]

(b) After year 1 the value is already 36\,400 > 26\,000.

Need 36\,400 \times 0.94^{n-1} < 26\,000

n - 1 > \frac{\ln(26000/36400)}{\ln 0.94} = 5.438

n = 7

A1[4]

Final answer

(a) \$28\,419
(b) 7 years

Question 25

[7 marks]

(a) Show that \displaystyle\sum_{r=1}^{n}\ln\left(1+\frac{2}{r}\right)=\ln\frac{(n+1)(n+2)}{2}.

[4]

(b) Determine the smallest n such that the sum is greater than 10.

[3]

Show solutionHide solution

(a) \ln\left(1+\frac2r\right)=\ln\frac{r+2}{r}=\ln(r+2)-\ln r.

Write out the terms: (\ln3-\ln1)+(\ln4-\ln2)+(\ln5-\ln3)+\dots+\bigl(\ln(n+1)-\ln(n-1)\bigr)+\bigl(\ln(n+2)-\ln n\bigr).

Every term cancels except \ln(n+1)+\ln(n+2)-\ln1-\ln2.

=\ln\frac{(n+1)(n+2)}{2} since \ln1=0.

A1AG

(b) \frac{(n+1)(n+2)}{2}>e^{10}=22026.47\ldots

n=208 gives 21945<22026; n=209 gives 22155>22026.

n=209.

Final answer

(a) shown
(b) n=209

Question 26

[15 marks]

Let f(x) = \ln(6x - 7) for x > \tfrac{7}{6}, and g(x) = 2\ln x - \ln c for x > 0, where c \in \mathbb{R}^{+}.

(a) State the equation of the vertical asymptote to the graph of y = g(x).

[1]

The two graphs meet at two distinct points.

(b) (i) Show that, at each point of intersection, x^{2} - 6cx + 7c = 0.

[3]

(ii) Hence show that 9c^{2} - 7c > 0.

[2]

(iii) Determine the range of possible values of c.

[4]

Denote the two x-coordinates of intersection by p and q, where p < q.

(c) For the case c = 7, find the exact value of q - p, writing it in the form a\sqrt{b} where a, b \in \mathbb{Z}^{+}.

[5]

Show solutionHide solution

(a) x = 0

A1[1]

(b)(i) \ln(6x - 7) = 2\ln x - \ln c

\ln\left[c(6x - 7)\right] = \ln x^{2} \implies 6cx - 7c = x^{2}

M1A1

x^{2} - 6cx + 7c = 0

AG[3]

(ii) \Delta > 0: 36c^{2} - 28c > 0

9c^{2} - 7c > 0

AGA1[2]

(iii) c(9c - 7) > 0 and c > 0, so 9c - 7 > 0

Both roots exceed \tfrac76: at x = \tfrac76, x^{2} - 6cx + 7c = \tfrac{49}{36} > 0, and the vertex x = 3c > \tfrac76

c > \tfrac{7}{9}

A1A1[4]

(c) x^{2} - 42x + 49 = 0

x = \frac{42 \pm \sqrt{1764 - 196}}{2} = 21 \pm \frac{\sqrt{1568}}{2} \quad \textbf{M1} \textbf{A1}

\sqrt{1568} = 28\sqrt{2}, \quad x = 21 \pm 14\sqrt{2}

q - p = 28\sqrt{2}

A1[5]

Final answer

(a) x=0
(b)(iii) c>\frac79
(c) q-p=28\sqrt2

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 The first digit of 2^n

G1 Worked example

[11 marks]

In this question \log means \log_{10}. For a number x\ge0, write x=k+f where k\in\mathbb{Z} and 0\le f<1; then f is the fractional part of x. The leading digit of a positive number is its first non-zero digit.

(a) Let N>0 and let f be the fractional part of \log N. Explain why N has leading digit d exactly when \log d\le f<\log(d+1).

[2]

(b) Given that \log7=0.845098\ldots, find the number of digits in 7^{20} and its leading digit.

[3]

(c) For n\in\mathbb{Z}^+, let f_n and g_n be the fractional parts of n\log2 and n\log5. Show that f_n+g_n=1.

[2]

(d) Hence show that if 2^n and 5^n have the same leading digit, then that digit is 3. Find the least n for which this happens.

[4]

Show solutionHide solution

(a) N=10^{k+f}=10^f\times10^k with 1\le10^f<10, so 10^f is N with its decimal point moved: it has the same leading digit as N

The leading digit is d exactly when d\le10^f<d+1, that is (since \log is increasing) when \log d\le f<\log(d+1)

(b) \log7^{20}=20\log7=16.90196\ldots

k=16, so 7^{20}=10^{0.90196\ldots}\times10^{16} has 17 digits

10^{0.90196\ldots}=7.979\ldots, so the leading digit is 7

Note: as a check, \log7=0.8451\le0.90196<0.9031=\log8.

(c) n\log2+n\log5=n\log10=n, a whole number

2^n has no factor 5, so it is not a power of 10 and f_n\ne0. With n\log2=k+f_n, n\log5=(n-k-1)+(1-f_n) where 0<1-f_n<1, so g_n=1-f_n and f_n+g_n=1

R1AG

(d) same leading digit d: by (a) and (c), \log d\le f_n<\log(d+1) and \log d\le1-f_n<\log(d+1)

adding the two left-hand inequalities and the two right-hand ones: 2\log d\le1<2\log(d+1), so d^2\le10<(d+1)^2

d=3

f_n must lie in [\log3,1-\log3]=[0.4771\ldots,0.5228\ldots]; f_1,\dots,f_4=0.301,\ 0.602,\ 0.903,\ 0.204 and f_5=0.505: n=5 (2^5=32, 5^5=3125)

Final answer

(b) 17 digits, leading digit 7
(d) n=5 (32 and 3125)

G1 Investigation

[13 marks]

For n\in\mathbb{Z}^+, let f_n be the fractional part of n\log2, as in the worked example.

(a) Use your GDC to write down the leading digits of 2^1,2^2,\dots,2^{20}. How many of these are 1?

[2]

(b) Suppose that 2^n has leading digit 1. Show that neither 2^{n+1} nor 2^{n+2} has leading digit 1, and that one of 2^{n+3} and 2^{n+4} does.

[4]

(c) Hence explain why, in a long list of consecutive powers of 2, the proportion with leading digit 1 is between about \frac14 and \frac13.

[2]

(d) Show that 2^n has leading digit 1 exactly when 0\le f_n<\log2. Write down the interval that contains f_n when 2^n has leading digit 7.

[2]

(e) You may assume that, as N increases, the proportion of the values f_1,f_2,\dots,f_N that lie in an interval [a,b)\subseteq[0,1) tends to b-a. Find the long-run proportion of powers of 2 with leading digit 1, and with leading digit 7. Of 2^1,2^2,\dots,2^{1000}, exactly 301 have leading digit 1 and 56 have leading digit 7. Comment.

[3]

Show solutionHide solution

(a) 2,4,8,1,3,6,1,2,5,1,\ 2,4,8,1,3,6,1,2,5,1

six of them are 1 (at n=4,7,10,14,17,20)

(b) write 2^n=m\times10^k with 1\le m<2

2^{n+1}=2m\times10^k with 2\le2m<4, and 2^{n+2}=4m\times10^k with 4\le4m<8: leading digits 2 or 3, then 4 to 7, never 1

8\le8m<16. If 8m\ge10, then 2^{n+3}=0.8m\times10^{k+1} with 1\le0.8m<1.6: leading digit 1

if 8\le8m<10, then 2^{n+4}=1.6m\times10^{k+1} with 1.6\le1.6m<2: leading digit 1

(c) by (b), after a power with leading digit 1 the next one comes 3 or 4 steps later

so a long run of N consecutive powers contains between about \frac N4 and \frac N3 of them

(d) 2^n=10^{f_n}\times10^k, so by the worked example the leading digit is 1 exactly when \log1\le f_n<\log2, that is 0\le f_n<\log2

leading digit 7: \log7\le f_n<\log8, that is [0.8451\ldots,0.9031\ldots)

(e) digit 1: \log2-\log1=\log2=0.301\ldots, which lies between \frac14 and \frac13, as (c) requires

digit 7: \log8-\log7=\log\frac87=0.0580\ldots

predicted counts for n\le1000: about 301 and about 58, against 301 and 56; the model fits well, and the share falls steadily from digit 1 to digit 9

Final answer

(a) 2,4,8,1,3,6,1,2,5,1,2,4,8,1,3,6,1,2,5,1; six are 1
(d) [\log7,\log8)
(e) \log2\approx0.301; \log\frac87\approx0.0580 (about 301 and 58 expected)

G2 Logarithms from square roots

G2 Worked example

[9 marks]

For x>0 and k\in\mathbb{N}, let s_k(x)=2^k\left(x^{1/2^k}-1\right). You may assume that \dfrac{e^h-1}{h}\to1 as h\to0.

(a) Use your GDC to find s_4(3) and s_{12}(3). Compare them with \ln3.

[2]

(b) For x\ne1, let h=\dfrac{\ln x}{2^k}. Show that s_k(x)=\ln x\times\dfrac{e^h-1}{h}, and hence that s_k(x)\to\ln x as k\to\infty.

[3]

(c) Show that \dfrac{x^{1/2^k}-1}{10^{1/2^k}-1}\to\log_{10}x as k\to\infty.

[2]

(d) Use k=10 in (c) to estimate \log_{10}2. Compare your estimate with the value from your GDC.

[2]

Show solutionHide solution

(a) s_4(3)=1.13720\ldots and s_{12}(3)=1.09875\ldots

\ln3=1.09861\ldots: the errors are about 0.039 and 0.00015, so more square roots give a better estimate

(b) x^{1/2^k}=e^{(\ln x)/2^k}=e^h

2^k=\dfrac{\ln x}{h}, so s_k(x)=\dfrac{\ln x}{h}\left(e^h-1\right)=\ln x\times\dfrac{e^h-1}{h}

as k\to\infty, h\to0, so the fraction tends to 1 and s_k(x)\to\ln x

(c) the factor 2^k cancels: \dfrac{x^{1/2^k}-1}{10^{1/2^k}-1}=\dfrac{s_k(x)}{s_k(10)}

\to\dfrac{\ln x}{\ln10}=\log_{10}x (change of base)

(d) \dfrac{2^{1/1024}-1}{10^{1/1024}-1}=0.300793\ldots

GDC: \log_{10}2=0.301029\ldots; the estimate is correct to 3 significant figures, with an error of about 2.4\times10^{-4}

Final answer

(a) 1.13720\ldots, 1.09875\ldots (\ln3=1.09861\ldots)
(d) 0.300793\ldots (\log_{10}2=0.301029\ldots)

G2 Investigation

[12 marks]

For k\in\mathbb{N}, let a_k=2^{1/2^k} and s_k=2^k(a_k-1), so that s_k\to\ln2, as in the worked example. Let e_k=s_k-\ln2.

(a) Use your GDC to find s_k and e_k for k=0,1,\dots,6. Comment on the values of \dfrac{e_{k+1}}{e_k}.

[3]

(b) Show that a_k=a_{k+1}^{\,2}, and hence that s_k-s_{k+1}=\dfrac{s_{k+1}^{\,2}}{2^{k+2}}.

[3]

(c) Deduce that s_k>s_{k+1} for every k. By considering s_k-s_{k+1} when k is large, explain why each error e_{k+1} is close to half of e_k.

[3]

(d) Use your answer to (c) to explain why t_k=2s_{k+1}-s_k should be a much better estimate of \ln2 than s_{k+1}. Find t_5 and its error.

[3]

Show solutionHide solution

(a) s_0,\dots,s_6=1,\ 0.828427,\ 0.756828,\ 0.724062,\ 0.708381,\ 0.700709,\ 0.696914

e_0,\dots,e_6=0.306853,\ 0.135280,\ 0.063681,\ 0.030915,\ 0.015233,\ 0.007562,\ 0.003767

\dfrac{e_{k+1}}{e_k}=0.441,\ 0.471,\ 0.485,\ 0.493,\ 0.496,\ 0.498: the ratios approach \frac12, so each error is about half the one before

(b) a_{k+1}^{\,2}=\left(2^{1/2^{k+1}}\right)^2=2^{2/2^{k+1}}=2^{1/2^k}=a_k

s_k=2^k\left(a_{k+1}^{\,2}-1\right)=2^k(a_{k+1}-1)(a_{k+1}+1)

a_{k+1}-1=\dfrac{s_{k+1}}{2^{k+1}}, so s_k=\dfrac{s_{k+1}}2\left(\dfrac{s_{k+1}}{2^{k+1}}+2\right)=s_{k+1}+\dfrac{s_{k+1}^{\,2}}{2^{k+2}}

A1AG

(c) a_{k+1}>1, so s_{k+1}>0 and s_k-s_{k+1}=\dfrac{s_{k+1}^{\,2}}{2^{k+2}}>0

for large k, s_{k+1}\approx\ln2, so d_k=s_k-s_{k+1}\approx\dfrac{(\ln2)^2}{2^{k+2}}: each difference is about half of the one before

the differences telescope: e_k=d_k+d_{k+1}+d_{k+2}+\dots\approx\dfrac{(\ln2)^2}{2^{k+2}}\left(1+\frac12+\frac14+\dots\right)=\dfrac{(\ln2)^2}{2^{k+1}}, which halves when k increases by 1

(d) e_k-e_{k+1}=s_k-s_{k+1} and e_k\approx2e_{k+1}, so e_{k+1}\approx s_k-s_{k+1}

\ln2=s_{k+1}-e_{k+1}\approx s_{k+1}-(s_k-s_{k+1})=2s_{k+1}-s_k: the main part of the error is removed

t_5=2s_6-s_5=0.693120\ldots; error -2.7\times10^{-5}, against e_6=3.8\times10^{-3}

Final answer

(a) ratios 0.441,\dots,0.498, approaching \frac12
(d) t_5=0.693120\ldots, error about -2.7\times10^{-5}

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 2 teaches exponentials and logarithms from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book