Free worked solutions · 2026 Edition

IB Maths AA SL 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 SL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

Chapter 2 of 12
22 practice questions
4 Going further
Free, no sign-up

Start with Basics All 12 chapters

Basics

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

Question 1

[4 marks]

Find the exact value of

\frac{(1.5\times 10^{4})\times(8\times 10^{-2})}{4\times 10^{-6}},

giving your answer in standard form.

[4]

Show solutionHide solution

numerator: (1.5\times 8)\times 10^{4-2}=12\times 10^{2}

=1.2\times 10^{3}

\dfrac{1.2\times 10^{3}}{4\times 10^{-6}}=0.3\times 10^{3-(-6)}=0.3\times 10^{9}

3\times 10^{8}

Note: accept the steps in any order.

Final answer

3\times10^8

Question 2

[6 marks]

In this question all letters stand for non-zero numbers.

Simplify each expression fully, giving your answer with positive exponents only.

(a) \left(2x^{3}y^{-1}\right)^{4}

[2]

(b) \dfrac{6a^{-2}b^{5}}{9a^{3}b^{-1}}

[2]

(c) \left(5k^{-1}\right)^{-2}

[2]

Show solutionHide solution

(a) 2^{4}x^{12}y^{-4}

=\dfrac{16x^{12}}{y^{4}}

(b) \dfrac{2}{3}a^{-2-3}b^{5-(-1)}=\dfrac{2}{3}a^{-5}b^{6}

=\dfrac{2b^{6}}{3a^{5}}

(c) 5^{-2}k^{2}

=\dfrac{k^{2}}{25}

Note: award M1 for correct use of the power or quotient law on at least two of the factors.

Final answer

(a) 16x^{12}/y^4
(b) 2b^6/(3a^5)
(c) k^2/25.

Question 3

[6 marks]

Solve each equation, giving exact answers.

(a) \log_{2} x=5

[1]

(b) \log x=-3

[1]

(c) \ln x=2

[1]

(d) Find x when \log_{5} x=0

[1]

(e) \log_{x} 81=4, where x>0, x\ne 1

[1]

(f) 4^{x}=\dfrac{1}{64}

[1]

Show solutionHide solution

(a) x=2^{5}=32

(b) x=10^{-3}=0.001

(c) x=e^{2}

(d) x=5^{0}=1

(e) x^{4}=81 and x>0, so x=3

(f) x=\log_{4}\dfrac{1}{64}=-3, since 4^{-3}=\dfrac{1}{64}

Note: in (c) do not accept a decimal approximation only, since an exact answer is required.

Final answer

(a) 32
(b) 1/1000
(c) e^2
(d) 1
(e) 3
(f) -3.

Question 4

[6 marks]

Using a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^{m}, find the exact value of each of the following.

(a) 1000^{\frac{2}{3}}

[2]

(b) 16^{\frac{5}{4}}

[2]

(c) \left(\dfrac{8}{125}\right)^{-\frac{2}{3}}

[2]

Show solutionHide solution

(a) \left(\sqrt[3]{1000}\right)^{2}=10^{2}

=100

(b) \left(\sqrt[4]{16}\right)^{5}=2^{5}

=32

(c) \left(\dfrac{125}{8}\right)^{\frac{2}{3}}=\left(\dfrac{5}{2}\right)^{2}

=\dfrac{25}{4}

Note: award M1 for either inverting the fraction or taking the cube root correctly.

Final answer

(a) 100
(b) 32
(c) 25/4.

Question 5

[8 marks]

Solve the following equations.

(a) \log_{2}x+\log_{2}6=5

[2]

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

[4]

(c) \ln x+\ln 4=2\ln 6

[2]

Show solutionHide solution

(a) \log_{2}(6x)=5\Rightarrow 6x=32

x=\dfrac{16}{3}

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

\dfrac{4x+1}{x-2}=9

4x+1=9x-18\Rightarrow x=\dfrac{19}{5}

x-2=\dfrac{9}{5}>0 and 4x+1>0, so both logs are defined and x=\dfrac{19}{5} is valid

(c) \ln(4x)=\ln 36

x=9

Final answer

(a) 16/3
(b) 19/5
(c) 9.

Question 6

[7 marks]

Solve the following equations without using logarithms.

(a) 2^x=32

[1]

(b) 3^{x-1}=\dfrac{1}{9}

[2]

(c) 5^{2x}=\sqrt{5}

[2]

(d) 10^{x+2}=0.001

[2]

Show solutionHide solution

(a) 2^x=2^5 \Rightarrow x=5

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

x-1=-2 \Rightarrow x=-1

(c) 5^{2x}=5^{\frac{1}{2}}

2x=\dfrac{1}{2} \Rightarrow x=\dfrac{1}{4}

(d) 10^{x+2}=10^{-3}

x+2=-3 \Rightarrow x=-5

Final answer

(a) 5
(b) -1
(c) 1/4
(d) -5.

Question 7

[5 marks]

Answer this question without a calculator.

(a) Rosa invests 2500 euros at a nominal annual rate of 8\%, compounded half-yearly. Find the value of her investment after 1 year.

[2]

(b) Find how much more this is than the value after 1 year if the 8\% were compounded yearly.

[1]

(c) Sami invests 10\,000 euros at a nominal annual rate of 12\%, compounded quarterly. Find the value of his investment after 6 months.

[2]

Show solutionHide solution

(a) 2500\times 1.04^2=2500\times 1.0816

=2704 euros

(b) 2704-2500\times 1.08=2704-2700=4 euros

(c) 2 quarters at 3\%: 10\,000\times 1.03^2

=10\,609 euros

Final answer

(a) 2704 euros
(b) 4 euros
(c) 10 609 euros.

Tutorial

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

Question 8

[6 marks]

Given that \log_{3}7=k, find expressions in terms of k for

(a) \log_{3}63;

[2]

(b) \log_{3}\!\left(\dfrac{7}{81}\right);

[2]

(c) \log_{3}\sqrt{21}.

[2]

Show solutionHide solution

(a) 63=9\times7, so \log_3 63=\log_3 9+\log_3 7.

[1]

=2+k.

[1]

(b) \log_3\dfrac{7}{81}=\log_3 7-\log_3 3^4.

[1]

=k-4.

[1]

(c) \log_3\sqrt{21}=\dfrac12\log_3(3\times7)=\dfrac12\left(\log_3 3+\log_3 7\right).

[1]

=\dfrac{1+k}{2}.

[1]

Final answer

(a) 2+k
(b) k-4
(c) (1+k)/2.

Question 9

[7 marks]

(a) Solve \log_{4}x+\log_{16}x=3.

[3]

(b) Solve \log_{x}2=\log_{4}x, where x>0, x\neq1. Give your answers in exact form.

[4]

Show solutionHide solution

(a) \log_{16} x=\dfrac{\log_4 x}{\log_4 16}=\dfrac12\log_4 x.

[1]

So \dfrac32\log_4 x=3, giving \log_4 x=2.

[1]

x=4^2=16.

[1]

(b) Let t=\log_2 x. Then \log_x2=\dfrac1t and \log_4x=\dfrac t2.

[2]

\dfrac1t=\dfrac t2, so t^2=2 and t=\pm\sqrt2.

[1]

x=2^{\sqrt2} or x=2^{-\sqrt2}.

[1]

Final answer

(a) 16
(b) 2^{\sqrt2} or 2^{-\sqrt2}.

Question 10

[8 marks]

Solve exactly

(a) 25^{x}-6\times5^{x}+5=0;

[4]

(b) e^{2x}-7e^{x}-18=0.

[4]

Show solutionHide solution

(a) Let u=5^x; then 25^x=u^2 and u^2-6u+5=0.

[1]

(u-1)(u-5)=0, so u=1 or u=5.

[1]

5^x=1 gives x=0.

[1]

5^x=5 gives x=1.

[1]

(b) Let u=e^x; then u^2-7u-18=0.

[1]

(u-9)(u+2)=0, so u=9 or u=-2.

[1]

e^x>0 for every real x, so u=-2 gives no solution.

[1]

e^x=9, so x=\ln9.

[1]

Final answer

(a) 0,1
(b) \ln9.

Question 11

[10 marks]

Consider f(x)=10-2^{\,x+1}.

(a) To obtain y=f(x) from y=2^x, three transformation steps are applied. Describe them in order.

[3]

(b) State the horizontal asymptote and the range.

[2]

(c) Solve f(x)=-6.

[2]

(d) Using your answers to parts (b) and (c), and the value of f(0), draw a labelled sketch of the curve. Include its asymptote, its y-intercept and the point from part (c).

[3]

Show solutionHide solution

(a) Translation by \begin{pmatrix}-1\\0\end{pmatrix} (one unit left): 2^x\to2^{x+1}.

[1]

Reflection in the x-axis: 2^{x+1}\to-2^{x+1}.

[1]

Translation by \begin{pmatrix}0\\10\end{pmatrix} (ten units up): \to10-2^{x+1}.

[1]

(Equally valid: vertical stretch with scale factor 2, then reflection in the x-axis, then translation up 10, since 2^{x+1}=2\times2^x.)

(b) The asymptote of y=2^x is y=0; after the sequence it becomes y=10.

[1]

2^{x+1}>0, so the range is f(x)<10.

[1]

(c) 10-2^{x+1}=-6, so 2^{x+1}=16=2^4.

[1]

x+1=4, so x=3.

[1]

(d) f(0)=10-2^1=8, so the y-intercept is (0,8).

[1]

The sketch shows the dashed asymptote y=10 (the curve approaches it from below as x\to-\infty), the curve through (0,8) and (3,-6), decreasing without bound as x increases.

[2]

Final answer

(a) Translation by (-1,0), reflection in the x-axis, translation by (0,10). (b) y=10, range y<10. (c) x=3. (d) Decreasing through (0,8) and (3,-6), approaching y=10 to the left.

Question 12

[7 marks]

Let f(x)=\log_{2}(x+3)-1 for x>-3.

(a) Find an expression for f^{-1}(x).

[3]

(b) Write down both the domain of f^{-1} and its range.

[2]

(c) Solve f^{-1}(x)=5.

[2]

Show solutionHide solution

(a) Write y=\log_2(x+3)-1 and swap x and y: x=\log_2(y+3)-1.

[1]

x+1=\log_2(y+3), so y+3=2^{x+1}.

[1]

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

[1]

(b) Domain of f^{-1} = range of f = all real x.

[1]

Range of f^{-1} = domain of f: y>-3.

[1]

(c) 2^{x+1}-3=5, so 2^{x+1}=8=2^3.

[1]

x=2.

[1]

Final answer

(a) 2^{x+1}-3
(b) domain \mathbb R, range y>-3
(c) 2.

Question 13

[9 marks]

For this question, f(x)=\ln(2x-1), while g(x)=2\ln(x-1).

(a) Write down the largest possible domain of f and of g.

[2]

(b) Show that any x-coordinate where the two curves meet satisfies x^{2}-4x+2=0.

[3]

(c) Hence find the exact x-coordinate of the point where the graphs intersect.

[2]

(d) Find the exact y-coordinate of this point, giving your answer in the form \ln k.

[2]

Show solutionHide solution

(a) f: 2x-1>0, so x>\dfrac12.

[1]

g: x-1>0, so x>1.

[1]

(b) 2\ln(x-1)=\ln(x-1)^2.

[1]

Equate the arguments: 2x-1=(x-1)^2.

[1]

2x-1=x^2-2x+1, so x^2-4x+2=0. [1]

(c) x=\dfrac{4\pm\sqrt{16-8}}{2}=2\pm\sqrt2.

[1]

2-\sqrt2\approx0.59<1 is outside the domain of g, so x=2+\sqrt2.

[1]

(d) y=f(2+\sqrt2)=\ln\!\big(2(2+\sqrt2)-1\big).

[1]

=\ln\!\left(3+2\sqrt2\right).

[1]

Final answer

(a) x>1/2, x>1
(b) x^2-4x+2=0
(c) 2+\sqrt2
(d) \ln(3+2\sqrt2).

Question 14

[6 marks]

The mass of an antibiotic, C mg, in a patient's blood t hours after an injection is modelled by C(t)=C_{0}\,e^{-kt}, where C_{0}=150. The half-life of the antibiotic in the blood is 5.5 hours.

(a) Show that k=\dfrac{\ln2}{5.5}.

[2]

(b) Find the time taken for the mass to fall to 10 mg, giving your answer in hours to three significant figures.

[2]

(c) Find the mass remaining after 30 hours, giving your answer to three significant figures.

[2]

Show solutionHide solution

(a) After 5.5 hours half remains: 75=150e^{-5.5k}, so e^{-5.5k}=\dfrac12.

[1]

-5.5k=\ln\dfrac12=-\ln2, so k=\dfrac{\ln2}{5.5}. [1]

(b) 150e^{-kt}=10, so e^{-kt}=\dfrac1{15} and t=\dfrac{\ln15}{k}.

[1]

With k=0.12603\ldots: t=21.5 hours (from 21.49\ldots).

[1]

(c) C(30)=150e^{-30k}.

[1]

=3.42 mg (from 3.4207\ldots).

[1]

Final answer

(a) k=\ln2/5.5
(b) 21.5 h
(c) 3.42 mg.

Question 15

[6 marks]

A can of drink is taken out of a fridge and left on a table in a warm room. The temperature of the drink, T\,{}^{\circ}C, t minutes after it is taken out is modelled by T(t)=24-19e^{-0.035t}.

(a) Write down the temperature of the drink at the moment it is taken out of the fridge.

[1]

(b) Find the time at which the drink reaches 15\,{}^{\circ}C, giving your answer in minutes to three significant figures.

[3]

(c) Find the temperature of the drink after 1 hour, to three significant figures.

[1]

(d) Write down the temperature of the room.

[1]

Show solutionHide solution

(a) T(0)=24-19=5, so 5\,{}^{\circ}C.

[1]

(b) 24-19e^{-0.035t}=15, so 19e^{-0.035t}=9.

[1]

e^{-0.035t}=\dfrac{9}{19}, so -0.035t=\ln\dfrac{9}{19}.

[1]

t=\dfrac{\ln(19/9)}{0.035}=21.3 minutes (from 21.35\ldots).

[1]

(c) T(60)=24-19e^{-2.1}=21.7\,{}^{\circ}C (from 21.67\ldots).

[1]

(d) e^{-0.035t}\to0, so T\to24: the room temperature is 24\,{}^{\circ}C.

[1]

Final answer

(a) 5\,{}^{\circ}C
(b) 21.3 min
(c) 21.7\,{}^{\circ}C
(d) 24\,{}^{\circ}C.

Question 16

[6 marks]

(a) Eli invests \$9000 at a nominal rate of r\% per year, with quarterly compounding. After 6 years the investment is worth \$11\,500. Find r, correct to three significant figures.

[3]

(b) Fay invests \$6500 at a nominal annual rate of 5.4\% compounded monthly. Find the least number of complete years after which her investment is worth more than \$15\,000.

[3]

Show solutionHide solution

(a) 9000\left(1 + \dfrac{r}{400}\right)^{24} = 11\,500.

[1]

1 + \dfrac{r}{400} = \left(\dfrac{11\,500}{9000}\right)^{1/24} = 1.010265\ldots (or: GDC solver with N = 24, PV = -9000, FV = 11\,500, P/Y = C/Y = 4).

[1]

r = 4.11 (3 s.f.).

[1]

(b) 6500\left(1 + \dfrac{5.4}{1200}\right)^{12t} > 15\,000.

[1]

GDC: t = 15.5\ldots (after 15 years: \$14\,584.88 < 15\,000; after 16 years: \$15\,392.25 > 15\,000).

[1]

Least number of complete years: 16.

[1]

Final answer

(a) 4.11\%
(b) 16 years.

Mastery

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

Question 17

[5 marks]

Solve \left(\log_{2}x\right)^{2}=\log_{2}\!\left(x^{5}\right)-4, where x>0.

Show solutionHide solution

\log_2\left(x^5\right)=5\log_2x; let t=\log_2x.

[1]

t^2=5t-4, so t^2-5t+4=0.

[1]

(t-1)(t-4)=0, so t=1 or t=4.

[1]

x=2^1=2.

[1]

x=2^4=16.

[1]

Final answer

x=2 or 16.

Question 18

[7 marks]

Solve \log_{3}(2x+1)+\log_{3}(x-1)=1+\log_{3}x. Give your answer in exact form and justify your choice of solution.

Show solutionHide solution

Left side: \log_3\big((2x+1)(x-1)\big).

[1]

Right side: 1=\log_3 3, so 1+\log_3x=\log_3(3x).

[1]

Equate the arguments: (2x+1)(x-1)=3x.

[1]

2x^2-x-1=3x, so 2x^2-4x-1=0.

[1]

x=\dfrac{4\pm\sqrt{16+8}}{4}.

[1]

x=1\pm\dfrac{\sqrt6}{2}.

[1]

\log_3(x-1) needs x>1, so reject 1-\dfrac{\sqrt6}2 (it is less than 1); the solution is x=1+\dfrac{\sqrt6}{2}.

[1]

Final answer

x=1+\sqrt6/2.

Question 19

[7 marks]

The mass M of a radioactive sample is modelled by M=M_{0}e^{-kt}, where t is the time in hours and k>0. After 8 hours, 36\% of the initial mass has decayed.

(a) Show that k=\dfrac14\ln\dfrac54.

[3]

(b) Show that the half-life of the sample is \dfrac{4\ln2}{\ln(5/4)} hours.

[2]

(c) Show that after 12 hours, 51.2\% of the initial mass remains.

[2]

Show solutionHide solution

(a) 64\% remains: e^{-8k}=0.64=\dfrac{16}{25}.

[1]

-8k=\ln\dfrac{16}{25}=-\ln\dfrac{25}{16}, so k=\dfrac18\ln\dfrac{25}{16}.

[1]

\ln\dfrac{25}{16}=\ln\left(\dfrac54\right)^2=2\ln\dfrac54, so k=\dfrac14\ln\dfrac54. [1]

(b) Half-life: e^{-kt}=\dfrac12, so kt=\ln2.

[1]

t=\dfrac{\ln2}{k}=\dfrac{4\ln2}{\ln(5/4)}. [1]

(c) e^{-12k}=\left(e^{-8k}\right)^{3/2}.

[1]

=0.64^{3/2}=0.8^3=0.512, i.e. 51.2\%. [1]

Final answer

(a) k=\frac14\ln(5/4)
(b) 4\ln2/\ln(5/4) h
(c) 51.2\%.

Question 20

[8 marks]

(a) Solve 2^{x}+2^{\,3-x}=9.

[5]

(b) Solve 2^{x}+2^{\,3-x}=7, giving your answers in exact form.

[3]

Show solutionHide solution

(a) Let u=2^x; then 2^{3-x}=\dfrac{8}{u}, so u+\dfrac8u=9.

[1]

u^2-9u+8=0.

[1]

(u-1)(u-8)=0, so u=1 or u=8.

[1]

2^x=1 gives x=0.

[1]

2^x=8 gives x=3.

[1]

(b) u+\dfrac8u=7 gives u^2-7u+8=0, so u=\dfrac{7\pm\sqrt{17}}{2}.

[2]

Both roots are positive (since \sqrt{17}<7), so both give solutions: x=\log_2\!\left(\dfrac{7+\sqrt{17}}{2}\right) or x=\log_2\!\left(\dfrac{7-\sqrt{17}}{2}\right).

[1]

Final answer

(a) x=0 or 3
(b) x=\log_2((7+\sqrt{17})/2) or \log_2((7-\sqrt{17})/2).

Question 21

[7 marks]

Gwen invests P dollars. For the first 4 years it earns a nominal annual rate of 3.6\% compounded monthly. It is then moved to another account paying a nominal annual rate of 4.2\% compounded quarterly for a further 3 years. At the end of the 7 years it is worth \$17\,800.

(a) Find P, to the nearest dollar.

[4]

(b) Find the total interest earned over the 7 years.

[1]

(c) Find the single nominal annual rate, compounded monthly for all 7 years, that would give the same final value from the same P. Give your answer to three significant figures.

[2]

Show solutionHide solution

(a) Stage 1 factor: \left(1 + \dfrac{0.036}{12}\right)^{48} = 1.003^{48}.

[1]

Stage 2 factor: \left(1 + \dfrac{0.042}{4}\right)^{12} = 1.0105^{12}.

[1]

P \times 1.003^{48} \times 1.0105^{12} = 17\,800.

[1]

P = \dfrac{17\,800}{1.003^{48} \times 1.0105^{12}} = 13\,600.0\ldots \approx \$13\,600.

[1]

(b) 17\,800-P\approx4200 dollars.

[1]

(c) P\left(1 + \dfrac{r}{1200}\right)^{84} = 17\,800 \Rightarrow 1 + \dfrac{r}{1200} = \left(\dfrac{17\,800}{P}\right)^{1/84}.

[1]

r = 3.85 (3 s.f.).

[1]

Final answer

(a) 13 600 dollars
(b) 4200 dollars
(c) 3.85\%.

Question 22

[15 marks]

A culture of yeast cells is grown in a laboratory. For the first 6 hours the number of yeast cells is modelled by N(t)=500\times1.4^{\,t}, where t is the time in hours since the culture was started. At t=6 a fungicide is added. For t>6 the number of yeast cells is modelled by

N(t)=N(6)\,e^{-0.35(t-6)}.

(a) Find N(6), to the nearest integer.

[2]

(b) Find the two times at which the number of yeast cells is 2000, each correct to three significant figures.

[4]

(c) Find the length of time, in hours to three significant figures, for which the number of yeast cells is greater than 2000.

[2]

(d) Find the time at which the number of yeast cells returns to its initial value of 500, correct to three significant figures.

[3]

(e) For t>6, find the time, in hours to three significant figures, for the number of yeast cells to halve.

[2]

(f) For t>6, show that the number of yeast cells decreases by the same percentage each hour, and find this percentage to three significant figures.

[2]

Show solutionHide solution

(a) N(6)=500\times1.4^6.

[1]

=3764.768\ldots=3765.

[1]

(b) First time (growth phase): 500\times1.4^t=2000, so 1.4^t=4.

[1]

t=\dfrac{\ln4}{\ln1.4}=4.12 hours.

[1]

Second time (decay phase): 3764.768\,e^{-0.35(t-6)}=2000, so e^{-0.35(t-6)}=0.53124123\ldots.

[1]

t-6=\dfrac{-\ln0.53124123}{0.35}=1.807\ldots, so t=7.81 hours.

[1]

(c) Time above 2000: 7.807254453\ldots-4.120085434\ldots.

[1]

=3.69 hours.

[1]

(d) 3764.768\,e^{-0.35(t-6)}=500, so e^{-0.35(t-6)}=0.13281\ldots.

[1]

-0.35(t-6)=\ln0.13281\ldots=-2.018833419\ldots, so t-6=5.768\ldots.

[1]

t=11.8 hours.

[1]

(e) Half: e^{-0.35\tau}=\dfrac12 where \tau=t-6.

[1]

\tau=\dfrac{\ln2}{0.35}=1.98 hours.

[1]

(f) \dfrac{N(t+1)}{N(t)}=\dfrac{e^{-0.35(t-5)}}{e^{-0.35(t-6)}}=e^{-0.35}, which does not depend on t.

[1]

e^{-0.35}=0.7047\ldots, so the number decreases by 29.5\% each hour.

[1]

Final answer

(a) 3765
(b) 4.12, 7.81 h
(c) 3.69 h
(d) 11.8 h
(e) 1.98 h
(f) 29.5\%.

Going further

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

G1 Worked example

[5 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\ge1 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 digit count and leading digit of 7^{20}.

[3]

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<0.90196\ldots<\log8.

Final answer

(a) N=10^f10^k, so d\le10^f<d+1. (b) 17 digits, leading digit 7.

G1 Investigation

[7 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 list the leading digits of 2^1,2^2,\dots,2^{20}. How many are 1?

[2]

(b) 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]

(c) 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) 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 [\log7,\log8)

(c) digit 1: \log2-\log1=\log2=0.301\ldots

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 ones. (b) [\log7,\log8). (c) \log2\approx0.301, \log(8/7)\approx0.0580; about 301 and 58 predicted. Observed 301 and 56 are close.

G2 Worked example

[7 marks]

For x>0 and k\in\mathbb{N}, let s_k(x)=2^k\left(x^{1/2^k}-1\right). You may use this supplied fact: when h is close to zero, (e^h-1)/h approaches 1.

(a) Use your GDC to find s_4(3) and s_{12}(3). Give both estimates to 5 decimal places and 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]

Show solutionHide solution

(a) s_4(3)\approx1.13721 and s_{12}(3)\approx1.09876

\ln3=1.09861\ldots: the errors are about 0.0386 and 0.000147, 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)

Final answer

(a) 1.13721, 1.09876; \ln3=1.09861. (b) As k\to\infty, h\to0 and the supplied ratio approaches 1. (c) \ln x/\ln10=\log_{10}x.

G2 Investigation

[9 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, to 6 decimal places. 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) You may assume that each error is close to half the preceding error. Explain why t_k=2s_{k+1}-s_k should be a much better estimate of \ln2 than s_{k+1}. Find t_5 to 6 decimal places, and its error to 3 significant figures.

[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) 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 -2.73\times10^{-5}, against e_6=3.8\times10^{-3}

Final answer

(a) s_k: 1, 0.828427, 0.756828, 0.724062, 0.708381, 0.700709, 0.696914; errors approximately halve. (b) s_k-s_{k+1}=s_{k+1}^2/2^{k+2}. (c) t_5=0.693120, error -2.73\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 12.

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 SL

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