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Exponents & Logs

Practice exam-style IB Math AA questions for Exponents & Logs, aligned with the syllabus and grouped by topic.

Paper
Difficulty
Status
Level
Question 1
SL • Paper 1
Easy
Non Calculator

A particle has mass 4.8×10−23 kg4.8\times 10^{-23}\,\text{kg}. A sample contains 2.5×10182.5\times 10^{18} identical particles.

A

Find the mass of the sample, in kg, giving your answer in the form a×10ka\times 10^k, where 1≤a<101\le a<10 and k∈Zk\in\mathbb{Z}.

[2]
Write your answer here...
B

Another sample has mass 3.6×10−2 kg3.6\times 10^{-2}\,\text{kg}. Find how many times as large this mass is as the mass found in part (a). Give your answer in standard form.

[3]
Write your answer here...

0

Question 2
SL • Paper 1
Easy
Non Calculator

Let x>0x>0 and

E=(2x−3)2(8x5)4x−2E=\frac{(2x^{-3})^2(8x^5)}{4x^{-2}}
A

Simplify EE in the form axbax^b, where a,b∈Za,b\in\mathbb{Z}.

[3]
Write your answer here...
B

Hence solve E=24E=24.

[2]
Write your answer here...

0

Question 3
SL • Paper 2
Easy
Calculator Permitted

Let AA and BB be positive real numbers such that log⁡A=2.37\log A=2.37 and ln⁡B=1.18\ln B=1.18, where log⁡\log denotes the logarithm to base 1010.

A

Find the value of AA.

[2]
Write your answer here...
B

Find the value of BB.

[2]
Write your answer here...
C

Find log⁡(A2B)\log\left(\dfrac{A^2}{\sqrt{B}}\right).

[2]
Write your answer here...

0

Question 4
SL • Paper 1
Medium
Non Calculator

Let a=log⁡102a=\log_{10}2 and b=log⁡105b=\log_{10}5.

A

Show that a+b=1a+b=1.

[2]
Write your answer here...
B

Write log⁡1040\log_{10}40 in terms of aa only.

[2]
Write your answer here...
C

Find log⁡4100\log_4 100 in terms of aa.

[2]
Write your answer here...

0

Question 5
SL • Paper 1
Medium
Non Calculator

Let x>0x>0.

A

Express 2log⁡3x−log⁡342\log_3 x-\log_3 4 as a single logarithm.

[2]
Write your answer here...
B

Hence solve 2log⁡3x−log⁡34=22\log_3 x-\log_3 4=2.

[3]
Write your answer here...

0

Question 6
SL • Paper 1
Medium
Non Calculator

Let x>0x>0 and

F=(x3/2)4x2xF=\frac{(x^{3/2})^4}{x^2\sqrt{x}}
A

Simplify FF as a single power of xx.

[3]
Write your answer here...
B

Given that F=128F=128, find xx.

[2]
Write your answer here...

0

Question 7
HL • Paper 1
Medium
Non Calculator

Let a>0a>0, a≠1a\ne1 and x>0x>0.

A

Show that

log⁡a2x+log⁡ax=52log⁡ax\log_{a^2}x+\log_{\sqrt a}x=\frac{5}{2}\log_a x
[3]
Write your answer here...
B

Hence solve log⁡9x+log⁡3x=10\log_9x+\log_{\sqrt3}x=10.

[2]
Write your answer here...

0

Question 8
SL • Paper 2
Medium
Calculator Permitted

A single bacterium has mass 7.40×10−13 g7.40\times 10^{-13}\ \text{g}. A sample of these bacteria has total mass 0.0360 g0.0360\ \text{g}. The number of bacteria in the sample is denoted by N0N_0. After the sample is placed in a nutrient solution, the number of bacteria is modelled by

N(t)=N02t25N(t)=N_0 2^{\frac{t}{25}}

where tt is the time in minutes.

A

Find N0N_0, giving your answer in the form a×10ka\times 10^k, where 1≤a<101\leq a<10 and k∈Zk\in\mathbb{Z}.

[2]
Write your answer here...
B

Determine the time taken for the number of bacteria to reach 1.00×10121.00\times 10^{12}.

[3]
Write your answer here...

0

Question 9
SL • Paper 2
Medium
Calculator Permitted

For x>0x>0, define

f(x)=(16x8)342x12f(x)=\frac{(16x^8)^{\frac{3}{4}}}{2x^{\frac{1}{2}}}
A

Simplify f(x)f(x) in the form axpax^p, where a,p∈Ra,p\in\mathbb{R}.

[3]
Write your answer here...
B

Solve f(x)=75f(x)=75.

[2]
Write your answer here...

0

Question 10
SL • Paper 2
Medium
Calculator Permitted

Consider the equation

log⁡3(x−1)+log⁡3(x+5)=2\log_3(x-1)+\log_3(x+5)=2

where x>1x>1.

A

Show that this equation can be written as x2+4x−14=0x^2+4x-14=0.

[2]
Write your answer here...
B

Hence solve the original equation, giving the exact value of xx.

[3]
Write your answer here...
C

Find log⁡3(x+5)\log_3(x+5) for this value of xx, giving your answer to 3 significant figures.

[1]
Write your answer here...

0

Question 11
SL • Paper 2
Medium
Calculator Permitted

A hot drink is placed in a room. The difference, DD, between the temperature of the drink and the room temperature is modelled by

D(t)=48(0.82)tD(t)=48(0.82)^t

where tt is the time in hours and DD is measured in degrees Celsius.

A

Find D(3)D(3).

[2]
Write your answer here...
B

Determine the time when D(t)=10D(t)=10.

[3]
Write your answer here...
C

State the first whole number of hours after which D(t)D(t) is less than 1010.

[1]
Write your answer here...

0

Question 12
HL • Paper 2
Medium
Calculator Permitted

The amount, AA, of a radioactive substance remaining after tt hours is modelled by

A(t)=Ce−ktA(t)=Ce^{-kt}

where CC and kk are positive constants. Initially there are 8.00×10−3 mol8.00\times 10^{-3}\ \text{mol} of the substance. After 1212 hours there are 2.30×10−3 mol2.30\times 10^{-3}\ \text{mol} remaining.

A

Write down the value of CC.

[1]
Write your answer here...
B

Find the value of kk.

[2]
Write your answer here...
C

Determine the time when 5.00×10−4 mol5.00\times 10^{-4}\ \text{mol} remains.

[2]
Write your answer here...
D

Find A(30)A(30), giving your answer in scientific notation.

[1]
Write your answer here...

0

Question 13
SL • Paper 1
Medium
Non Calculator
A

Solve the equation 32x−1=5x+13^{2x-1}=5^{x+1}, giving your answer in exact form.

[4]
Write your answer here...

0

Question 14
HL • Paper 1
Medium
Non Calculator

Consider the equation

4x−9⋅2x+18=04^x-9\cdot2^x+18=0
A

Let t=2xt=2^x. Write the equation as a quadratic equation in tt.

[2]
Write your answer here...
B

Hence solve the original equation, giving your answers in exact form.

[3]
Write your answer here...

0

Question 15
HL • Paper 1
Medium
Non Calculator

Let x>0x>0, x≠1x\ne1. Consider the equation

log⁡x16+log⁡4x=3\log_x16+\log_4x=3
A

By using the substitution t=log⁡4xt=\log_4x, show that t2−3t+2=0t^2-3t+2=0.

[3]
Write your answer here...
B

Hence find all possible values of xx.

[3]
Write your answer here...

0

Question 16
HL • Paper 1
Medium
Non Calculator

Let a,b,c>0a,b,c>0, with a≠1a\ne1 and b≠1b\ne1.

A

Prove that

log⁡ab⋅log⁡bc=log⁡ac\log_a b\cdot\log_b c=\log_a c
[3]
Write your answer here...
B

Hence evaluate

log⁡23⋅log⁡35⋅log⁡57⋅log⁡732\log_2 3\cdot\log_3 5\cdot\log_5 7\cdot\log_7 32
[2]
Write your answer here...

0

Question 17
HL • Paper 1
Medium
Non Calculator

Consider the inequality

(12)x2−3x<4x−1\left(\frac12\right)^{x^2-3x}<4^{x-1}
A

Rewrite the inequality using powers of 22 only.

[2]
Write your answer here...
B

Hence solve the inequality.

[3]
Write your answer here...

0

Question 18
SL • Paper 2
Medium
Calculator Permitted

The sound level, LL, in decibels, is given by

L=10log⁡(II0)L=10\log\left(\frac{I}{I_0}\right)

where II is the sound intensity and I0=1.00×10−12 W m−2I_0=1.00\times 10^{-12}\ \text{W m}^{-2}.

A

Find LL when I=3.60×10−7 W m−2I=3.60\times 10^{-7}\ \text{W m}^{-2}.

[2]
Write your answer here...
B

Sound A is 18.0 dB18.0\ \text{dB} louder than sound B. Find the ratio of the intensity of sound A to the intensity of sound B.

[2]
Write your answer here...
C

Two identical machines each produce intensity II. Find the increase in sound level when both machines operate together instead of one machine operating alone.

[1]
Write your answer here...

0

Question 19
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

2x+e−x=42^x+e^{-x}=4
A

Solve the equation, giving both solutions correct to three significant figures.

[3]
Write your answer here...
B

Solve 2x=e−x2^x=e^{-x}.

[1]
Write your answer here...
C

Hence, or otherwise, solve the inequality 2x+e−x<42^x+e^{-x}<4.

[2]
Write your answer here...

0

Question 20
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

log⁡2x+log⁡x16=5\log_2 x+\log_x 16=5

where x>0x>0 and x≠1x\neq 1.

A

By letting u=log⁡2xu=\log_2 x, show that u2−5u+4=0u^2-5u+4=0.

[3]
Write your answer here...
B

Hence solve the original equation.

[2]
Write your answer here...

0

Question 21
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

log⁡4(3x−1)=log⁡2(x+1)−1\log_4(3x-1)=\log_2(x+1)-1
A

Write the equation using logarithms to base 22 only.

[1]
Write your answer here...
B

Show that the equation can be written as x2−10x+5=0x^2-10x+5=0.

[2]
Write your answer here...
C

Hence solve the original equation.

[2]
Write your answer here...

0

Question 22
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

(9x)12+31−x=7(9^x)^{\frac{1}{2}}+3^{1-x}=7
A

By letting u=3xu=3^x, show that u2−7u+3=0u^2-7u+3=0.

[3]
Write your answer here...
B

Hence solve the original equation, giving your answers correct to three significant figures.

[3]
Write your answer here...

0

Question 23
SL • Paper 1
Medium
Non Calculator

A thin filter sheet is made from identical fibres. Each fibre has mass 3.2×10−14 kg3.2\times 10^{-14}\ \text{kg}. One layer of the sheet contains 2.5×1092.5\times 10^9 fibres, and each sheet contains 8.0×1028.0\times 10^2 layers.

A
I.

Find the mass of one layer, giving your answer in the form a×10ka\times 10^k, where 1≤a<101\leq a<10 and k∈Zk\in\mathbb{Z}.

[2]
Write your answer here...
II.

Find the mass of one sheet, giving your answer in the form a×10ka\times 10^k, where 1≤a<101\leq a<10 and k∈Zk\in\mathbb{Z}.

[2]
Write your answer here...
B

container holds 1.28×103 kg1.28\times 10^3\ \text{kg} of these sheets. Find the number of sheets in the container.

[2]
Write your answer here...
C
I.

Each sheet has area 4.0×10−3 m24.0\times 10^{-3}\ \text{m}^2. Find the total area of the sheets in the container.

[1]
Write your answer here...
II.

Find the fractional increase of this area over 75 m275\ \text{m}^2.

[2]
Write your answer here...

0

Question 24
SL • Paper 1
Medium
Non Calculator

Let x>0x>0 and define

E=(27x6)233x12E=\frac{(27x^6)^{\frac{2}{3}}}{3x^{\frac{1}{2}}}
A
I.

Simplify EE in the form axpax^p, where a,p∈Ra,p\in\mathbb{R}.

[3]
Write your answer here...
II.

Hence solve E=242E=24\sqrt{2}. If you did not obtain E=3x72E=3x^{\frac{7}{2}}, use this expression.

[2]
Write your answer here...
B
I.

Show that E2=9x7E^2=9x^7. If you did not obtain E=3x72E=3x^{\frac{7}{2}}, use this expression.

[2]
Write your answer here...
II.

Hence solve E=12xE=\dfrac{12}{x}, giving your answer in exact form.

[3]
Write your answer here...

0

Question 25
SL • Paper 1
Medium
Non Calculator

Let a=log⁡25a=\log_2 5.

A
AI.

Express log⁡220\log_2 20 in terms of aa.

[1]
Write your answer here...
AII.

Express log⁡45\log_4 5 in terms of aa.

[1]
Write your answer here...
AIII.

Express log⁡102\log_{10}2 in terms of aa.

[2]
Write your answer here...
B
BI.

Solve 5x=85^x=8, giving your answer in terms of aa.

[2]
Write your answer here...
BII.

Solve 20y=5020^y=50, giving your answer in terms of aa.

[2]
Write your answer here...
C

Show that 0<a<30<a<3.

[2]
Write your answer here...

0

Question 26
SL • Paper 1
Medium
Non Calculator

The pH of a solution is defined by

pH=−log⁡H\text{pH}=-\log H

where HH is the hydrogen ion concentration in mol dm−3\text{mol dm}^{-3} and log⁡\log denotes logarithm to base 1010.

A
I.

Find the pH when H=10−3H=10^{-3}.

[1]
Write your answer here...
II.

solution has pH 7.47.4. Express its value of HH in the form 10q10^q.

[2]
Write your answer here...
B

Solution B has hydrogen ion concentration 100100 times that of solution A, where solution A has pH 33. Find the pH of solution B.

[2]
Write your answer here...
C
I.

Equal volumes of two solutions, with concentrations 10−310^{-3} and 10−510^{-5}, are mixed. Find the concentration of the mixture in scientific notation.

[2]
Write your answer here...
II.

Find the pH of the mixture, leaving your answer in exact logarithmic form.

[2]
Write your answer here...

0

Question 27
HL • Paper 1
Medium
Non Calculator

Let x>0x>0. Consider the equation

log⁡2(8x)log⁡2(x2)=15\log_2(8x)\log_2\left(\frac{x}{2}\right)=15
A

By using the substitution u=log⁡2xu=\log_2x, show that this equation may be written as u2+2u−18=0u^2+2u-18=0.

[4]
Write your answer here...
B

Hence solve the original equation, giving your answers in exact form.

[3]
Write your answer here...

0

Question 28
SL • Paper 2
Medium
Calculator Permitted

A space probe transmits a signal with power P0=7.20×1015 WP_0=7.20\times10^{15}\ \text{W}. At a distance rr metres from the probe, the intensity II of the signal is modelled by

I=P4πr2I=\frac{P}{4\pi r^2}

where PP is the transmitted power in watts.

A
I.

Find the intensity of the signal when r=3.00×1011r=3.00\times10^{11} metres.

[2]
Write your answer here...
II.

The receiving station can detect signals with intensity at least 5.00×10−10 W m−25.00\times10^{-10}\ \text{W m}^{-2}. Determine the greatest distance from which the signal can be detected, assuming P=P0P=P_0. Give your answer in scientific notation.

[2]
Write your answer here...
B

Treat tt as continuous. Determine the initial intensity at this distance and the time at which the intensity reaches the detection limit. Hence state the cutoff time after which the signal is undetectable.

[3]
Write your answer here...
C

For a different receiving station, the detection limit is reduced by a factor of 100100. Determine how this changes the greatest detectable distance, assuming the same transmitted power P0P_0.

[3]
Write your answer here...

0

Question 29
SL • Paper 2
Medium
Calculator Permitted

The pH of an aqueous solution is defined by

pH=−log⁡[H+]\text{pH}=-\log[\text{H}^+]

where [H+][\text{H}^+] is the hydrogen ion concentration in mol dm−3\text{mol dm}^{-3} and log⁡\log denotes logarithm to base 1010.

A
I.

Solution A has [H+]=3.20×10−4 mol dm−3[\text{H}^+]=3.20\times10^{-4}\ \text{mol dm}^{-3}. Find its pH.

[2]
Write your answer here...
II.

Solution B has pH 4.204.20. Find [H+][\text{H}^+] for solution B.

[1]
Write your answer here...
B
I.

25.0 cm325.0\ \text{cm}^3 sample of solution A is diluted to a total volume of 500 cm3500\ \text{cm}^3. Find the new hydrogen ion concentration.

[2]
Write your answer here...
II.

Find the pH of the diluted solution.

[2]
Write your answer here...
C

Water is added to 40.0 cm340.0\ \text{cm}^3 of solution A until the pH is 5.005.00. Determine the volume of water added.

[3]
Write your answer here...

0

Question 30
SL • Paper 2
Medium
Calculator Permitted

For x>0x>0, define

f(x)=(81x12)143x−12f(x)=\frac{(81x^{12})^{\frac14}}{3x^{-\frac12}}
A
I.

Show that f(x)=x72f(x)=x^{\frac72}.

[2]
Write your answer here...
II.

Find f(1.8)f(1.8).

[1]
Write your answer here...
B

Solve f(x)=90f(x)=90, giving your answer correct to three significant figures.

[3]
Write your answer here...
C

Solve log⁡2(f(x))−log⁡2(x+1)=3\log_2(f(x))-\log_2(x+1)=3 for x>0x>0.

[3]
Write your answer here...

0

Question 31
SL • Paper 2
Medium
Calculator Permitted

The magnitude MM of an earthquake is modelled by

M=log⁡(AA0)M=\log\left(\frac{A}{A_0}\right)

where AA is the amplitude of the seismic wave, A0A_0 is a reference amplitude, and log⁡\log denotes logarithm to base 1010.

A
I.

An earthquake has magnitude 5.805.80. Find AA0\frac{A}{A_0}.

[1]
Write your answer here...
II.

second earthquake has magnitude 4.504.50. Find the ratio of the amplitude of the first earthquake to the amplitude of the second.

[2]
Write your answer here...
B

Two seismic waves arrive at the same station at the same time. Their individual magnitudes are 5.805.80 and 4.504.50, and their amplitudes add. Find the magnitude of the combined wave.

[3]
Write your answer here...
C

After the first earthquake, the amplitude at a station is modelled by A(t)=A1e−0.18tA(t)=A_1e^{-0.18t}, where tt is measured in minutes and A1A_1 is the initial amplitude corresponding to magnitude 5.805.80. Determine the time at which the magnitude reaches 3.203.20, and hence state when it is less than 3.203.20.

[4]
Write your answer here...

0

Question 32
HL • Paper 2
Medium
Calculator Permitted

Let a>0a>0, a≠1a\neq 1. It is given that

log⁡a12≈2.54andlog⁡a5≈1.65\log_a12\approx2.54\quad\text{and}\quad \log_a5\approx1.65
A

Find log⁡a(14425)\log_a\left(\dfrac{144}{25}\right).

[2]
Write your answer here...
B

Find the value of aa.

[2]
Write your answer here...
C

Solve a2x−17ax+60=0a^{2x}-17a^x+60=0.

[3]
Write your answer here...

0

Question 33
HL • Paper 3
Medium
Calculator Permitted

A data archive stores images. Each uncompressed image has size SS bytes. The archive uses binary prefixes, so 1 GiB=2301\ \text{GiB}=2^{30} bytes. The number of images stored after nn months is modelled by

Nn=N0(1.18)nN_n=N_0(1.18)^n

nn

NnN_n

Storage [bytes]

Capacity [bytes]

0

2.50×1052.50 \times 10^{5}

1.20×10121.20 \times 10^{12}

8.00×10128.00 \times 10^{12}

1

2.95×1052.95 \times 10^{5}

1.42×10121.42 \times 10^{12}

8.00×10128.00 \times 10^{12}

2

3.48×1053.48 \times 10^{5}

1.67×10121.67 \times 10^{12}

8.00×10128.00 \times 10^{12}

3

4.11×1054.11 \times 10^{5}

1.97×10121.97 \times 10^{12}

8.00×10128.00 \times 10^{12}

4

4.85×1054.85 \times 10^{5}

2.33×10122.33 \times 10^{12}

8.00×10128.00 \times 10^{12}

5

5.72×1055.72 \times 10^{5}

2.75×10122.75 \times 10^{12}

8.00×10128.00 \times 10^{12}

6

6.75×1056.75 \times 10^{5}

3.24×10123.24 \times 10^{12}

8.00×10128.00 \times 10^{12}

7

7.96×1057.96 \times 10^{5}

3.82×10123.82 \times 10^{12}

8.00×10128.00 \times 10^{12}

8

9.40×1059.40 \times 10^{5}

4.51×10124.51 \times 10^{12}

8.00×10128.00 \times 10^{12}

9

1.11×1061.11 \times 10^{6}

5.32×10125.32 \times 10^{12}

8.00×10128.00 \times 10^{12}

10

1.31×1061.31 \times 10^{6}

6.28×10126.28 \times 10^{12}

8.00×10128.00 \times 10^{12}

11

1.54×1061.54 \times 10^{6}

7.41×10127.41 \times 10^{12}

8.00×10128.00 \times 10^{12}

12

1.82×1061.82 \times 10^{6}

8.75×10128.75 \times 10^{12}

8.00×10128.00 \times 10^{12}

13

2.15×1062.15 \times 10^{6}

1.03×10131.03 \times 10^{13}

8.00×10128.00 \times 10^{12}

14

2.54×1062.54 \times 10^{6}

1.22×10131.22 \times 10^{13}

8.00×10128.00 \times 10^{12}

15

2.99×1062.99 \times 10^{6}

1.44×10131.44 \times 10^{13}

8.00×10128.00 \times 10^{12}

16

3.53×1063.53 \times 10^{6}

1.70×10131.70 \times 10^{13}

8.00×10128.00 \times 10^{12}

17

4.17×1064.17 \times 10^{6}

2.00×10132.00 \times 10^{13}

8.00×10128.00 \times 10^{12}

18

4.92×1064.92 \times 10^{6}

2.36×10132.36 \times 10^{13}

8.00×10128.00 \times 10^{12}

19

5.80×1065.80 \times 10^{6}

2.79×10132.79 \times 10^{13}

8.00×10128.00 \times 10^{12}

20

6.85×1066.85 \times 10^{6}

3.29×10133.29 \times 10^{13}

8.00×10128.00 \times 10^{12}

21

8.08×1068.08 \times 10^{6}

3.88×10133.88 \times 10^{13}

8.00×10128.00 \times 10^{12}

22

9.54×1069.54 \times 10^{6}

4.58×10134.58 \times 10^{13}

8.00×10128.00 \times 10^{12}

23

1.13×1071.13 \times 10^{7}

5.40×10135.40 \times 10^{13}

8.00×10128.00 \times 10^{12}

24

1.33×1071.33 \times 10^{7}

6.37×10136.37 \times 10^{13}

8.00×10128.00 \times 10^{12}

A
I.

Given S=4.80×106S=4.80\times10^6 and N0=2.50×105N_0=2.50\times10^5, find the initial storage required in bytes, in scientific notation.

[2]
Write your answer here...
II.

Convert this storage requirement to GiB.

[1]
Write your answer here...
B

The archive has capacity 8.00×10128.00\times10^{12} bytes.

I.

Form an inequality for the number of complete months before the capacity is exceeded.

[2]
Write your answer here...
II.

Determine the first month in which the capacity is exceeded.

[2]
Write your answer here...
C

Instead of allowing storage to grow by 18%18\% each month, the archive compresses all images at the end of each month by a fixed factor qq, where 0<q<10<q<1. Thus, for part (c), the modified model is

Nn=N0(1.18q)nN_n=N_0(1.18q)^n

The monthly multiplier is therefore 1.18q1.18q.

I.

Find the greatest value of qq that would keep the archive within capacity for 2424 complete months.

[3]
Write your answer here...
II.

Explain why the inequality sign is not reversed when solving for qq.

[1]
Write your answer here...

0

Question 34
HL • Paper 3
Medium
Calculator Permitted

A training app models the time TnT_n in minutes taken by a student to complete a puzzle on attempt nn by

Tn=A+Be−knT_n=A+Be^{-kn}

where AA is the limiting completion time and B,k>0B,k>0.

Completion time models decrease toward the limiting time A=6.
A

For one student, A=6A=6, T1=22T_1=22 and T4=14T_4=14.

I.

Show that e−3k=12e^{-3k}=\dfrac12.

[2]
Write your answer here...
II.

Find kk and BB.

[3]
Write your answer here...
B

Use the model found in part (a). If you did not obtain values for BB and kk, use B=20.2B=20.2 and k=0.231k=0.231.

I.

Find the first attempt number for which the completion time is less than 88 minutes.

[2]
Write your answer here...
II.

Explain why the model never predicts a completion time below 66 minutes.

[1]
Write your answer here...
C

second student has the same limiting time A=6A=6 and the same initial excess time as the first student, namely 16 min16\ \text{min}. For this student, use the model Tn=6+B2e−k2nT_n=6+B_2e^{-k_2n}, where k2=1.2kk_2=1.2k and B2e−k2=16 minB_2e^{-k_2}=16\ \text{min}.

I.

Find this student's predicted time on attempt 44.

[2]
Write your answer here...
II.

For the model Tn=A+Be−knT_n=A+Be^{-kn}, discuss whether increasing kk has a larger effect for small nn or for large nn. Hold AA and BB fixed and define the effect as the absolute change in TnT_n when kk is increased by 20%20\%.

[2]
Write your answer here...

0

Question 35
SL • Paper 1
Hard
Non Calculator

Let x>2x>2 and

L=log⁡3(x−2)+log⁡3(x+6)−log⁡38L=\log_3(x-2)+\log_3(x+6)-\log_3 8
A
I.

Express LL as a single logarithm.

[2]
Write your answer here...
II.

Show that the equation L=2L=2 can be written as x2+4x−84=0x^2+4x-84=0.

[2]
Write your answer here...
B

Hence solve L=2L=2, giving your answer in exact form.

[2]
Write your answer here...
C

Determine the number of solutions of the equation L=kL=k, where k∈Rk\in\mathbb{R}.

[4]
Write your answer here...

0

Question 36
SL • Paper 1
Hard
Non Calculator

The difference, DD, between the temperature of an object and the surrounding temperature is modelled by

D(t)=81(23)tD(t)=81\left(\frac{2}{3}\right)^t

where tt is the time in hours and D(t)D(t) is measured in degrees Celsius.

A
I.

Find D(2)D(2).

[2]
Write your answer here...
II.

Write down the value of D(t+1)D(t)\dfrac{D(t+1)}{D(t)}.

[1]
Write your answer here...
B
I.

Find the exact time when D(t)=16D(t)=16.

[2]
Write your answer here...
II.

Find the exact solution of the inequality D(t)<8D(t)<8.

[2]
Write your answer here...
C

Determine the least integer value of nn for which D(n)<1D(n)<1.

[3]
Write your answer here...

0

Question 37
HL • Paper 1
Hard
Non Calculator

A light source emits power P=3.2×1026 WP=3.2\times 10^{26}\ \text{W}. At a distance rr metres from the source, the intensity II is modelled by

I=P4πr2I=\frac{P}{4\pi r^2}
A
I.

Find r2r^2 when r=4.0×1011r=4.0\times 10^{11}. Give your answer in scientific notation.

[2]
Write your answer here...
II.

Find the intensity when r=4.0×1011r=4.0\times 10^{11}.

[2]
Write your answer here...
B

At what distance is the intensity equal to 5π\dfrac{5}{\pi}?

[3]
Write your answer here...
C

Show that multiplying the distance by 10n10^n divides the intensity by 102n10^{2n}.

[3]
Write your answer here...

0

Question 38
SL • Paper 2
Hard
Calculator Permitted

The concentration CC of a medicine in a patient’s bloodstream is modelled by

C(t)=18atC(t)=18a^t

where tt is the time in hours after the first dose, CC is measured in mg L−1\text{mg L}^{-1}, and 0<a<10<a<1. It is known that C(6)=7.50C(6)=7.50.

A
I.

Find the value of aa.

[2]
Write your answer here...
II.

Find the time taken for the concentration from the first dose to halve.

[2]
Write your answer here...
B

Determine the threshold-crossing time after the first dose, when the concentration reaches 2.50 mg L−12.50\ \text{mg L}^{-1}.

[3]
Write your answer here...
C
I.

second dose is given 88 hours after the first dose. This adds 10 mg L−110\ \text{mg L}^{-1} to the concentration immediately. Find the total concentration immediately after the second dose.

[2]
Write your answer here...
II.

After the second dose, the total concentration is modelled by the sum of the remaining concentration from each dose. Assume that the concentration from the second dose decays with the same factor aa as the first dose. Determine the threshold-crossing time after the first dose, when the total concentration reaches 5.00 mg L−15.00\ \text{mg L}^{-1}. If you did not obtain a value for aa, use a=0.864a=0.864.

[2]
Write your answer here...

0

Question 39
SL • Paper 2
Hard
Calculator Permitted

A laboratory investigates the intensity II of light, in lux, after it passes through dd millimetres of tinted glass. The data are modelled by

I=AbdI=Ab^d

where A>0A>0 and 0<b<10<b<1.

d [mm]

I [lux]

0

150.0

1

132.3

2

116.6

3

102.9

4

90.7

5

80.0

6

70.6

7

62.3

8

54.9

9

48.4

10

42.7

11

37.7

12

33.2

A
I.

Use the data to find an exponential regression model for II in the form I=AbdI=Ab^d.

[2]
Write your answer here...
II.

Interpret the value of bb in the context of the model.

[2]
Write your answer here...
B

Use your model to estimate the thickness of glass needed to reduce the intensity to 40.040.0 lux.

[3]
Write your answer here...
C

second piece of glass of the same type is placed after a first piece of thickness d1d_1. The total thickness is d1+d2d_1+d_2. Explain why the model predicts that the percentage loss in intensity caused by the second piece depends only on d2d_2, not on d1d_1.

[4]
Write your answer here...

0

Question 40
HL • Paper 2
Hard
Calculator Permitted

Let a>1a>1. For real xx, define

F(x)=log⁡ax+log⁡x(a4)F(x)=\log_a x+\log_x(a^4)

wherever the expression is defined.

A
I.

State the restrictions on xx required for F(x)F(x) to be defined.

[1]
Write your answer here...
II.

Let u=log⁡axu=\log_a x. Express log⁡x(a4)\log_x(a^4) in terms of uu.

[1]
Write your answer here...
B

Show that solving F(x)=kF(x)=k is equivalent to solving u2−ku+4=0u^2-ku+4=0, where u=log⁡axu=\log_a x.

[3]
Write your answer here...
C

Hence solve F(x)=5F(x)=5, giving your answers in terms of aa.

[2]
Write your answer here...
D

Determine the values of kk for which F(x)=kF(x)=k has exactly two distinct solutions for xx.

[3]
Write your answer here...

0

Question 41
HL • Paper 2
Hard
Calculator Permitted

Let x>3x>3 and y>1y>1. Consider the system

log⁡2x+log⁡2y=5\log_2 x+\log_2 y=5 log⁡2(x−3)+log⁡2(y−1)=3\log_2(x-3)+\log_2(y-1)=3
A
I.

Show that xy=32xy=32.

[2]
Write your answer here...
II.

Show that x+3y=27x+3y=27.

[2]
Write your answer here...
B

Hence find all possible ordered pairs (x,y)(x,y).

[2]
Write your answer here...
C

For each ordered pair found in part (b), calculate log⁡2(x+3y)\log_2(x+3y) and comment on why the value is the same for both pairs.

[3]
Write your answer here...

0

Question 42
HL • Paper 3
Hard
Calculator Permitted

A sensor records the brightness BB of a lamp after it is switched on. The model used is

B(t)=Bmax(1−e−kt)B(t)=B_{\text{max}}\left(1-e^{-kt}\right)

where tt is the time in seconds, BmaxB_{\text{max}} is the limiting brightness and k>0k>0. The graph shows the brightness increasing towards a horizontal asymptote.

Exponential brightness curve approaching a limiting brightness of 80 units.
A
I.

Given that Bmax=80B_{\text{max}}=80 and B(12)=50B(12)=50, show that k=0.0817 s−1k=0.0817\ \text{s}^{-1}, correct to three significant figures.

[3]
Write your answer here...
II.

Using the unrounded value of kk obtained in part (a)(i), before rounding to three significant figures, find the time at which the brightness first reaches 7070.

[2]
Write your answer here...
B

second lamp follows the model C(t)=100(1−at)C(t)=100(1-a^t), where 0<a<10<a<1. It reaches 6060 units of brightness after 1515 seconds.

I.

Find the exact value of aa in the form m1/nm^{1/n}, where mm is a positive rational number and nn is a positive integer.

[2]
Write your answer here...
II.

Write C(t)C(t) in the form 100(1−e−λt)100(1-e^{-\lambda t}) and find λ\lambda.

[1]
Write your answer here...
C

For a general model D(t)=L(1−e−kt)D(t)=L(1-e^{-kt}), where LL is the limiting brightness, define TpT_p to be the time required to reach p%p\% of the limiting brightness, where 0<p<1000<p<100.

I.

Show that Tp=−1kln⁡(1−p100)T_p=-\dfrac{1}{k}\ln\left(1-\dfrac{p}{100}\right).

[3]
Write your answer here...
II.

Hence find the exact value of T90T50\dfrac{T_{90}}{T_{50}} and interpret this value in context.

[2]
Write your answer here...

0

Question 43
HL • Paper 3
Hard
Calculator Permitted

The energy EE released by a meteor impact is estimated from its diameter dd metres using

E=4.20×109d3E=4.20\times 10^{9}d^3

where EE is measured in joules. A logarithmic impact index II is defined by

I=log⁡(E1012)I=\log\left(\frac{E}{10^{12}}\right)

where log⁡\log denotes base 1010.

Straight-line graph of impact index I against log₁₀ d.
A
I.

Find EE when d=75d=75, giving your answer in scientific notation.

[2]
Write your answer here...
II.

Find the impact index for d=75d=75.

[2]
Write your answer here...
B

The index can be written in the form I=A+3log⁡dI=A+3\log d, where AA is a constant.

I.

Show that A=log⁡(4.20)−3A=\log(4.20)-3.

[2]
Write your answer here...
II.

Use this result to determine the diameter of a meteor with impact index 4.004.00.

[2]
Write your answer here...
C

Two meteors have diameters d1d_1 and d2d_2. Their impact indices satisfy I2−I1=1.50I_2-I_1=1.50.

I.

Show that d2d1=100.5\dfrac{d_2}{d_1}=10^{0.5} if the second meteor has the larger index.

[3]
Write your answer here...
II.

Hence determine the ratio of the released energies E2E1\dfrac{E_2}{E_1}, giving your answer in scientific notation.

[2]
Write your answer here...
III.

Explain why a straight-line graph is obtained when II is plotted against log⁡d\log d, and state its gradient.

[1]
Write your answer here...

0

Question 44
HL • Paper 3
Hard
Calculator Permitted

For a weak acid solution, the hydrogen ion concentration HH in moles per litre is related to its pH by

pH=−log⁡H\text{pH}=-\log H

A laboratory dilutes a solution by adding pure water. The concentration after nn equal dilution steps is modelled by

Hn=H0rnH_n=H_0r^n

where 0<r<10<r<1.

n

H [mol L^-1]

pH

0

3.20×10^-4

3.4949

1

1.60×10^-4

3.7959

2

8.00×10^-5

4.0969

3

4.00×10^-5

4.3979

4

2.00×10^-5

4.6990

5

1.00×10^-5

5.0000

6

5.00×10^-6

5.3010

7

2.50×10^-6

5.6021

8

1.25×10^-6

5.9031

9

6.25×10^-7

6.2041

A
I.

solution has H0=3.20×10−4H_0=3.20\times10^{-4}. Find its pH.

[2]
Write your answer here...
II.

After 55 dilution steps, H5=1.00×10−5H_5=1.00\times10^{-5}. Find rr.

[2]
Write your answer here...
B

Assume r=0.500r=0.500 for the rest of the question.

I.

Show that the pH after nn dilution steps is 3.49+nlog⁡23.49+n\log 2, correct to three significant figures in the constant term.

[2]
Write your answer here...
II.

Determine the smallest integer value of nn for which the pH exceeds 6.006.00.

[1]
Write your answer here...
C

different solution has pH P0P_0 before dilution. Each dilution step multiplies HH by the same factor rr.

I.

Prove that the increase in pH after one dilution step is −log⁡r-\log r.

[3]
Write your answer here...
II.

technician wants each step to increase the pH by exactly 0.2500.250. Find the required dilution factor rr.

[2]
Write your answer here...

0

Question 45
HL • Paper 3
Hard
Calculator Permitted

A capacitor discharges through a resistor. The voltage VV after time tt seconds is modelled by

V(t)=V0e−t/τV(t)=V_0e^{-t/\tau}

where V0V_0 is the initial voltage and τ\tau is a positive constant called the time constant.

Exponential voltage decay of a discharging capacitor with a 3.00 V threshold line.
A
I.

Given that V0=12.0V_0=12.0 and V(4)=7.20V(4)=7.20, find τ\tau.

[2]
Write your answer here...
II.

Using your value of τ\tau, find the time for the voltage to fall to 1.501.50 volts.

[2]
Write your answer here...
B

Engineers sometimes use base 22 half-life notation instead of base ee notation.

I.

Show that V(t)=V0(12)t/hV(t)=V_0\left(\dfrac12\right)^{t/h}, where h=τln⁡2h=\tau\ln2.

[3]
Write your answer here...
II.

Find hh for this capacitor.

[1]
Write your answer here...
C

safety circuit uses a fixed threshold of 3.003.00 volts. Define the activation time as the time at which the voltage falls to this threshold. The initial voltage is doubled, while τ\tau is unchanged.

I.

Find an expression for the increase in activation time caused by doubling V0V_0.

[3]
Write your answer here...
II.

Interpret this result in terms of half-life.

[1]
Write your answer here...

0

Question 46
HL • Paper 3
Hard
Calculator Permitted

Atmospheric pressure PP in kilopascals at height hh kilometres above sea level is modelled by

P(h)=101e−chP(h)=101e^{-ch}

where c>0c>0.

Atmospheric pressure decreases exponentially with altitude, with sea-level and summit points marked.
A
I.

At 2.402.40 km above sea level, the pressure is 75.075.0 kPa. Find cc.

[2]
Write your answer here...
II.

Using this model, find the pressure at 5.505.50 km.

[2]
Write your answer here...
B

Let R=P(h)101 kPaR=\dfrac{P(h)}{101\ \text{kPa}} be the pressure as a fraction of sea-level pressure.

I.

Show that h=−ln⁡Rch=-\dfrac{\ln R}{c}.

[2]
Write your answer here...
II.

Find the altitude at which the pressure is half the sea-level pressure.

[1]
Write your answer here...
C

The model may also be written as:

P(h)=101×10−ahP(h)=101\times10^{-ah}

where a>0a>0.

I.

Express aa in terms of cc.

[2]
Write your answer here...
II.

mountaineer incorrectly uses a=ca=c. Determine the percentage error in the predicted pressure at 6.006.00 km, using c=0.124c=0.124.

[3]
Write your answer here...
III.

State the mathematical reason for the error.

[1]
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0

Question 47
HL • Paper 3
Hard
Calculator Permitted

A biologist models the metabolic rate MM of an animal by a power law

M=kWpM=kW^p

where WW is the animal's mass in kilograms, and k,pk,p are positive constants. Taking logarithms gives a linear relationship.

The exact calibration data are (W,M)=(4,18)(W,M)=(4,18) and (W,M)=(64,144)(W,M)=(64,144). The graph displays the corresponding logarithmic coordinates (1,2.085)(1,2.085) and (3,3.585)(3,3.585) rounded to three decimal places.

Two calibration points on the log-log M–W relation.
A

For two animals, (W,M)=(4,18)(W,M)=(4,18) and (64,144)(64,144).

I.

Use the two data points to find pp exactly.

[3]
Write your answer here...
II.

Find kk exactly.

[2]
Write your answer here...
B

Use M=922W3/4M=\dfrac{9\sqrt2}{2}W^{3/4}.

I.

Find the mass of an animal with metabolic rate 300300 units.

[2]
Write your answer here...
II.

If mass is multiplied by 1616, determine the factor by which metabolic rate is multiplied.

[2]
Write your answer here...
C

Let X=logWX=\\log W and Y=log⁡MY=\log M, where the logarithm may be taken in any fixed base.

I.

Show that the graph of YY against XX is a straight line and state its gradient.

[3]
Write your answer here...
II.

Justify why the value of the gradient does not depend on the base of the logarithm used.

[2]
Write your answer here...

0

Question 48
HL • Paper 1
Hard
Non Calculator

Let a>0a>0, a≠1a\neq 1 and x>0x>0. Let y=log⁡axy=\log_a x.

A
I.

Show that, for r≠0r\neq 0, log⁡arx=yr\log_{a^r}x=\dfrac{y}{r}.

[3]
Write your answer here...
II.

Hence simplify P=log⁡a2x+log⁡a−1x+log⁡a13xP=\log_{a^2}x+\log_{a^{-1}}x+\log_{a^{\frac{1}{3}}}x in terms of yy.

[3]
Write your answer here...
B

It is given that P=10P=10 and a2x=32a^2x=32.

I.

Find yy. If you did not obtain P=5y2P=\dfrac{5y}{2}, use this expression.

[1]
Write your answer here...
II.

Find the exact values of aa and xx.

[3]
Write your answer here...
III.

Find log⁡xa\log_x a.

[2]
Write your answer here...

0

Question 49
HL • Paper 1
Hard
Non Calculator

Consider the equation

4x+4−x=1744^x+4^{-x}=\frac{17}{4}
A
I.

By letting u=2xu=2^x, show that the equation may be written as 4u4−17u2+4=04u^4-17u^2+4=0.

[3]
Write your answer here...
II.

Solve 4u4−17u2+4=04u^4-17u^2+4=0 for u>0u>0.

[3]
Write your answer here...
III.

Hence solve the original equation.

[2]
Write your answer here...
B

Determine all real values of kk for which 4x+4−x=k4^x+4^{-x}=k has at least one real solution.

[4]
Write your answer here...

0

Question 50
HL • Paper 1
Hard
Non Calculator

Let x>0x>0, x≠1x\neq 1. Consider the equation

log⁡x32+log⁡8x=143\log_x 32+\log_8 x=\frac{14}{3}
A
I.

Let u=log⁡2xu=\log_2 x. Express log⁡x32\log_x 32 and log⁡8x\log_8 x in terms of uu.

[3]
Write your answer here...
II.

Show that u2−14u+15=0u^2-14u+15=0.

[2]
Write your answer here...
B

Hence solve the original equation, giving your answers in exact form.

[4]
Write your answer here...
C

For x>1x>1, determine the least possible value of log⁡x32+log⁡8x\log_x 32+\log_8x.

[3]
Write your answer here...

0

Question 51
HL • Paper 1
Hard
Non Calculator

Let a>0a>0, a≠1a\neq 1, x>0x>0 and x≠1x\neq 1. Consider

log⁡ax+log⁡xa=c\log_a x+\log_x a=c
A
I.

Let u=log⁡axu=\log_a x. Show that the equation can be written as u+1u=cu+\dfrac{1}{u}=c.

[3]
Write your answer here...
II.

Find the possible values of uu when c=52c=\dfrac{5}{2}.

[2]
Write your answer here...
B

Given that a=16a=16 and c=52c=\dfrac{5}{2}, find all possible values of xx.

[2]
Write your answer here...
C

Assume now that a>1a>1 and x>ax>a. Determine all values of cc for which the equation has exactly one solution for xx.

[5]
Write your answer here...

0

Question 52
HL • Paper 1
Hard
Non Calculator

Let p>0p>0, p≠1p\neq 1. Consider the expression

Q=p2x−(p+1)px+pQ=p^{2x}-(p+1)p^x+p
A
I.

By letting t=pxt=p^x, factorize QQ in terms of tt.

[2]
Write your answer here...
II.

Hence solve Q=0Q=0 for xx.

[2]
Write your answer here...
B

Solve Q<0Q<0 for xx when p>1p>1.

[3]
Write your answer here...
C

Solve Q<0Q<0 for xx when 0<p<10<p<1.

[3]
Write your answer here...
D

Show that Q<0Q<0 has the same solution set for every permitted value of pp.

[2]
Write your answer here...

0

Question 53
HL • Paper 2
Hard
Calculator Permitted

Consider the function

f(x)=2x+3−x,x∈Rf(x)=2^x+3^{-x},\quad x\in\mathbb{R}
Graph of f(x)=2^x+3^-x with the line y=4.
A
I.

Use a GDC to solve f(x)=4f(x)=4.

[2]
Write your answer here...
II.

State the interval on which f(x)<4f(x)<4.

[1]
Write your answer here...
B

Find the minimum value of f(x)f(x) and the value of xx at which it occurs.

[4]
Write your answer here...
C

Determine the number of solutions of 2x+3−x=k2^x+3^{-x}=k for different real values of kk.

[3]
Write your answer here...

0

Question 54
HL • Paper 2
Hard
Calculator Permitted

Consider the equation

22x+5⋅2−x=72^{2x}+5\cdot2^{-x}=7
A
I.

Let u=2xu=2^x. Show that the equation can be written as u3−7u+5=0u^3-7u+5=0.

[3]
Write your answer here...
II.

Explain why only positive roots of this cubic are relevant.

[1]
Write your answer here...
B

Use a GDC to find the positive roots of u3−7u+5=0u^3-7u+5=0.

[2]
Write your answer here...
C

Hence solve 22x+5⋅2−x<72^{2x}+5\cdot2^{-x}<7.

[3]
Write your answer here...

0

Question 55
HL • Paper 2
Hard
Calculator Permitted

In a chemical experiment, a reaction rate RR is modelled by

R=Ae−BTR=Ae^{-\frac{B}{T}}

where TT is the temperature in kelvin, and AA and BB are positive constants. The rate is 1.80×10−51.80\times10^{-5} when T=290T=290, and 6.70×10−46.70\times10^{-4} when T=330T=330.

A
I.

Show that ln⁡R=ln⁡A−BT\ln R=\ln A-\frac{B}{T}.

[1]
Write your answer here...
II.

Use the two given rates to show that B≈8650B\approx8650.

[2]
Write your answer here...
B

Find the value of AA. Give your answer in scientific notation.

[4]
Write your answer here...
C

Determine the temperature at which the model predicts R=1.00×10−3R=1.00\times10^{-3}. If you did not obtain values for AA and BB, use A=1.62×108A=1.62\times10^8 and B=8650B=8650.

[4]
Write your answer here...

0

Question 56
HL • Paper 3
Hard
Calculator Permitted

The apparent brightness FF of a star is measured in watts per square metre. Astronomers define an apparent magnitude mm by

m=−2.5log⁡(FF0)m=-2.5\log\left(\frac{F}{F_0}\right)

where F0F_0 is a reference brightness.

Star

Apparent brightness FF [W m−2^{-2}]

Apparent magnitude mm

Star A

3.60×10−63.60\times10^{-6}

−5.00-5.00

Star B

3.60×10−73.60\times10^{-7}

−2.50-2.50

Reference

3.60×10−83.60\times10^{-8}

0.000.00

Star C

3.60×10−93.60\times10^{-9}

2.502.50

Star D

3.60×10−103.60\times10^{-10}

5.005.00

Star E

3.60×10−113.60\times10^{-11}

7.507.50

A
I.

Given F0=3.60×10−8F_0=3.60\times10^{-8} and F=9.00×10−10F=9.00\times10^{-10}, find mm.

[2]
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II.

Find FF when m=6.00m=6.00, giving your answer in scientific notation.

[2]
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B

Two stars have brightnesses F1F_1 and F2F_2, and magnitudes m1m_1 and m2m_2.

I.

Show that m2−m1=−2.5log⁡(F2F1)m_2-m_1=-2.5\log\left(\dfrac{F_2}{F_1}\right).

[3]
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II.

Hence find the brightness ratio F2:F1F_2:F_1 if star 22 is 3.003.00 magnitudes greater than star 11.

[2]
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C

cluster contains NN identical stars, each with brightness FF and magnitude mm. The total brightness is NFNF.

I.

Derive an expression for the magnitude MNM_N of the cluster in terms of the magnitude mm of one star and NN.

[3]
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II.

If each star has magnitude 8.008.00, find the least integer NN for which the cluster has magnitude less than 2.002.00.

[1]
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0

Question 57
HL • Paper 3
Hard
Calculator Permitted

In a chemical kinetics experiment, the rate constant KK is modelled by

K=Ae−E/(RT)K=Ae^{-E/(RT)}

where TT is the temperature in kelvin and A,E,RA,E,R are positive constants. Taking natural logarithms gives

ln⁡K=ln⁡A−ER(1T)\ln K=\ln A-\frac{E}{R}\left(\frac1T\right)

The graph shows experimental values of ln⁡K\ln K plotted against 1/T1/T.

Arrhenius plot of ln K against 1/T, with experimental points and a continuous decreasing best-fit line extrapolated to the y-intercept.
A

For one reaction, a line of best fit has equation y=−6200x+18.4y=-6200x+18.4, where y=ln⁡Ky=\ln K and x=1Tx=\dfrac1T.

I.

State the value of ln⁡A\ln A.

[1]
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II.

Find AA, giving your answer in scientific notation.

[2]
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III.

Given R=8.31R=8.31, find EE.

[1]
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B

Use the line of best fit from part (a).

I.

Estimate KK when T=350T=350 K.

[2]
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II.

Determine the temperature at which K=10.0K=10.0.

[2]
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C

student instead plots log⁡K\log K against 1/T1/T, using base 1010 logarithms.

I.

Suppose log⁡K\log K is plotted against 1/T1/T, using common logarithms. Find the gradient and intercept of this new line.

[3]
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II.

Explain why the same experimental data still lie approximately on a straight line.

[1]
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D

Let K1=K(T1)K_1=K(T_1) and K2=K(T2)K_2=K(T_2), and define q=K2K1q=\frac{K_2}{K_1}. The model predicts that increasing the temperature from T1T_1 to T2T_2 multiplies the rate constant by qq.

I.

Show that ln⁡q=ER(1T1−1T2)\ln q=\dfrac{E}{R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right).

[2]
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II.

Using E/R=6200E/R=6200, find qq when the temperature increases from T1=340 KT_1=340\ \text{K} to T2=360 KT_2=360\ \text{K}.

[1]
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0

Question 58
HL • Paper 3
Hard
Calculator Permitted

A data scientist studies first digits of numbers that are spread evenly on a logarithmic scale. In this model, the probability that the first digit is dd is

P(d)=log⁡(d+1)−log⁡dP(d)=\log(d+1)-\log d

for d=1,2,…,9d=1,2,\ldots,9, where log⁡\log denotes base 1010.

Bar chart of the first-digit probabilities P(d) for d = 1, 2, ..., 9.
A
I.

Show that P(d)=log⁡(1+1d)P(d)=\log\left(1+\dfrac1d\right).

[2]
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II.

Find P(1)P(1) and P(9)P(9) correct to three significant figures.

[2]
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B

Consider the sum of all nine probabilities.

I.

Prove that ∑d=19P(d)=1\sum_{d=1}^{9}P(d)=1.

[3]
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II.

Explain why this supports interpreting P(d)P(d) as a probability distribution.

[1]
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C

The model is generalised to leading two-digit numbers. For a leading two-digit number nn, where n=10,11,…,99n=10,11,\ldots,99, let its probability be

Q(n)=log⁡(n+1)−log⁡nQ(n)=\log(n+1)-\log n
I.

Prove that ∑n=1099Q(n)=1\sum_{n=10}^{99}Q(n)=1.

[3]
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II.

Determine the probability that the leading two-digit number is from 5050 to 5959 inclusive.

[2]
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0

Question 59
HL • Paper 3
Hard
Calculator Permitted

A transparent filter transmits a fraction TT of incoming light. Its absorbance AA is defined by

A=−log⁡TA=-\log T

where 0<T≤10<T\le1 and log⁡\log denotes base 1010. When filters are stacked, their transmission fractions multiply.

Part

Filter arrangement

Transmission data

Absorbance data

Other given values and condition

(a)(i)

One filter

T=0.180T=0.180

Not given

—

(a)(ii)

One filter

Not given

A=1.30A=1.30

—

(b)(i)

Two filters in series

T1T_1, T2T_2; Ttotal=T1T2T_{\mathrm{total}}=T_1T_2

A1A_1, A2A_2

—

(b)(ii)

Three identical filters in series

Ttotal=0.200%=0.00200T_{\mathrm{total}}=0.200\%=0.00200

Not given

—

(c)(i)

One filter

Not given

A=0.37A=0.37

—

(c)(ii)

nn identical filters in series

Not given

A=0.37A=0.37 each

Iin=6.50×10−3 W m−2I_{\mathrm{in}}=6.50\times10^{-3}\ \mathrm{W\,m^{-2}}; Ith=1.00×10−6 W m−2I_{\mathrm{th}}=1.00\times10^{-6}\ \mathrm{W\,m^{-2}}; activation when Itrans≤IthI_{\mathrm{trans}}\le I_{\mathrm{th}}

A
I.

filter transmits 18.0%18.0\% of the incoming light. Find its absorbance.

[2]
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II.

Find the transmission fraction of a filter with absorbance 1.301.30.

[2]
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B

Two filters have transmission fractions T1T_1 and T2T_2, and absorbances A1A_1 and A2A_2.

I.

Prove that the absorbance of the two filters stacked together is A1+A2A_1+A_2.

[3]
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II.

Three identical filters are stacked and transmit 0.200%0.200\% of the incoming light. Find the absorbance of one filter.

[1]
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C

manufacturer produces filters of absorbance 0.370.37 each. The incoming light intensity is 6.50×10−3 W m−26.50\times10^{-3}\ \text{W m}^{-2}. A detector activates when the transmitted intensity is less than or equal to IthI_{\mathrm{th}}, where Ith=1.00×10−6 W m−2I_{\mathrm{th}}=1.00\times10^{-6}\ \text{W m}^{-2}.

I.

Find the transmission fraction of one filter.

[1]
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II.

Determine the least number of filters needed for the detector to activate.

[3]
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III.

Explain why the inequality sign reverses when isolating nn in part (c)(ii).

[2]
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0

Question 60
HL • Paper 2
Hard
Calculator Permitted

For −2<x<5-2<x<5, define

f(x)=log⁡3(x+2)−log⁡3(5−x)f(x)=\log_3(x+2)-\log_3(5-x)
Graph of f(x) on -2<x<5 with y=2x-1.
A
I.

State the domain of ff.

[1]
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II.

Solve f(x)=0f(x)=0.

[1]
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B

Show that the inverse function is

f−1(y)=5⋅3y−23y+1f^{-1}(y)=\frac{5\cdot3^y-2}{3^y+1}
[3]
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C

Use a GDC to solve f(x)=2x−1f(x)=2x-1. Give all solutions in the domain.

[3]
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D

Explain why the answer x=5.00x=5.00 to three significant figures must be interpreted carefully.

[3]
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0

Counting Principles

Proofs