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Exponents & Logs

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

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Question 1
SL • Paper 1
Easy
Non Calculator
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

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]
B

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

[3]
Question 2
SL • Paper 1
Easy
Non Calculator
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]
B

E=24E=24.

[2]
Question 3
SL • Paper 2
Easy
Calculator Permitted
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

the value of AA.

[2]
B

the value of BB.

[2]
C

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

[2]
Question 4
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

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

A

a+b=1a+b=1.

[2]
B

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

[2]
C

log⁡4100\log_4 100 in terms of aa.

[2]

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Question 5
SL • Paper 1
Medium
Non Calculator
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]
B

2log⁡3x−log⁡34=22\log_3 x-\log_3 4=2.

[3]
Question 6
SL • Paper 1
Medium
Non Calculator
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]
B

Given that F=128F=128, xx.

[2]
Question 7
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let a>0a>0, a≠1a\ne1 and x>0x>0.

A

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

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

[2]
Question 8
SL • Paper 2
Medium
Calculator Permitted
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

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]
B

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

[3]

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Question 9
SL • Paper 2
Medium
Calculator Permitted
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]
B

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

[2]
Question 10
SL • Paper 2
Medium
Calculator Permitted
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

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

[2]
B

the original equation, giving the exact value of xx.

[3]
C

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

[1]
Question 11
SL • Paper 2
Medium
Calculator Permitted
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

D(3)D(3).

[2]
B

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

[3]
C

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

[1]
Question 12
HL • Paper 2
Medium
Calculator Permitted
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

the value of CC.

[1]
B

the value of kk.

[2]
C

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

[2]
D

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

[1]

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Question 13
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A

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

[4]
Question 14
HL • Paper 1
Medium
Non Calculator
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]
B

the original equation, giving your answers in exact form.

[3]
Question 15
HL • Paper 1
Medium
Non Calculator
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, t2−3t+2=0t^2-3t+2=0.

[3]
B

all possible values of xx.

[3]
Question 16
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

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

A

that

log⁡ab⋅log⁡bc=log⁡ac\log_a b\cdot\log_b c=\log_a c
[3]
B

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]

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Question 17
HL • Paper 1
Medium
Non Calculator
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]
B

the inequality.

[3]
Question 18
SL • Paper 2
Medium
Calculator Permitted
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

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

[2]
B

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

[2]
C

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

[1]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

2x+e−x=42^x+e^{-x}=4
A

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

[3]
B

2x=e−x2^x=e^{-x}.

[1]
C

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

[2]
Question 20
HL • Paper 2
Medium
Calculator Permitted
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, u2−5u+4=0u^2-5u+4=0.

[3]
B

the original equation.

[2]

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Question 21
HL • Paper 2
Medium
Calculator Permitted
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]
B

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

[2]
C

the original equation.

[2]
Question 22
HL • Paper 2
Medium
Calculator Permitted
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, u2−7u+3=0u^2-7u+3=0.

[3]
B

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

[3]
Question 23
SL • Paper 1
Medium
Non Calculator
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.

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]
II.

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]
B

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

[2]
C
I.

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

[1]
II.

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

[2]
Question 24
SL • Paper 1
Medium
Non Calculator
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]
II.

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

[2]
B
I.

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

[2]
II.

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

[3]

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Question 25
SL • Paper 1
Medium
Non Calculator
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]
AII.

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

[1]
AIII.

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

[2]
B
BI.

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

[2]
BII.

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

[2]
C

0<a<30<a<3.

[2]
Question 26
SL • Paper 1
Medium
Non Calculator
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.

the pH when H=10−3H=10^{-3}.

[1]
II.

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

[2]
B

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

[2]
C
I.

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

[2]
II.

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

[2]
Question 27
HL • Paper 1
Medium
Non Calculator
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, this equation may be written as u2+2u−18=0u^2+2u-18=0.

[4]
B

the original equation, giving your answers in exact form.

[3]
Question 28
SL • Paper 2
Medium
Calculator Permitted
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.

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

[2]
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}. the greatest distance from which the signal can be detected, assuming P=P0P=P_0. Give your answer in scientific notation.

[2]
B

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

[3]
C

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

[3]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
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}. its pH.

[2]
II.

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

[1]
B
I.

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

[2]
II.

the pH of the diluted solution.

[2]
C

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

[3]
Question 30
SL • Paper 2
Medium
Calculator Permitted
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.

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

[2]
II.

f(1.8)f(1.8).

[1]
B

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

[3]
C

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

[3]
Question 31
SL • Paper 2
Medium
Calculator Permitted
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. AA0\frac{A}{A_0}.

[1]
II.

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

[2]
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. the magnitude of the combined wave.

[3]
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. the time at which the magnitude reaches 3.203.20, and when it is less than 3.203.20.

[4]
Question 32
HL • Paper 2
Medium
Calculator Permitted
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

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

[2]
B

the value of aa.

[2]
C

a2x−17ax+60=0a^{2x}-17a^x+60=0.

[3]

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Question 33
HL • Paper 3
Medium
Calculator Permitted
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, the initial storage required in bytes, in scientific notation.

[2]
II.

Convert this storage requirement to GiB.

[1]
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]
II.

the first month in which the capacity is exceeded.

[2]
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.

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

[3]
II.

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

[1]
Question 34
HL • Paper 3
Medium
Calculator Permitted
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.

A

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

I.

e−3k=12e^{-3k}=\dfrac12.

[2]
II.

kk and BB.

[3]
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.

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

[2]
II.

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

[1]
C

A 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.

this student's predicted time on attempt 44.

[2]
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]
Question 35
SL • Paper 1
Hard
Non Calculator
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]
II.

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

[2]
B

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

[2]
C

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

[4]
Question 36
SL • Paper 1
Hard
Non Calculator
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.

D(2)D(2).

[2]
II.

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

[1]
B
I.

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

[2]
II.

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

[2]
C

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

[3]

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Question 37
HL • Paper 1
Hard
Non Calculator
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.

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

[2]
II.

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

[2]
B

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

[3]
C

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

[3]
Question 38
SL • Paper 2
Hard
Calculator Permitted
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.

the value of aa.

[2]
II.

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

[2]
B

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

[3]
C
I.

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

[2]
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. 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]
Question 39
SL • Paper 2
Hard
Calculator Permitted
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 an exponential regression model for II in the form I=AbdI=Ab^d.

[2]
II.

the value of bb in the context of the model.

[2]
B

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

[3]
C

A 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. 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]
Question 40
HL • Paper 2
Hard
Calculator Permitted
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.

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

[1]
II.

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

[1]
B

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]
C

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

[2]
D

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

[3]

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Question 41
HL • Paper 2
Hard
Calculator Permitted
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.

xy=32xy=32.

[2]
II.

x+3y=27x+3y=27.

[2]
B

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

[2]
C

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

[3]
Question 42
HL • Paper 3
Hard
Calculator Permitted
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.

A
I.

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

[3]
II.

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

[2]
B

A 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.

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]
II.

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

[1]
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.

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

[3]
II.

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

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
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.

A
I.

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

[2]
II.

the impact index for d=75d=75.

[2]
B

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

I.

A=log⁡(4.20)−3A=\log(4.20)-3.

[2]
II.

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

[2]
C

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

I.

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

[3]
II.

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

[2]
III.

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

[1]
Question 44
HL • Paper 3
Hard
Calculator Permitted
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.

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

[2]
II.

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

[2]
B

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

I.

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]
II.

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

[1]
C

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

I.

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

[3]
II.

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

[2]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
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.

A
I.

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

[2]
II.

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

[2]
B

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

I.

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

[3]
II.

hh for this capacitor.

[1]
C

A 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.

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

[3]
II.

this result in terms of half-life.

[1]
Question 46
HL • Paper 3
Hard
Calculator Permitted
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.

A
I.

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

[2]
II.

Using this model, the pressure at 5.505.50 km.

[2]
B

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

I.

h=−ln⁡Rch=-\dfrac{\ln R}{c}.

[2]
II.

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

[1]
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]
II.

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

[3]
III.

the mathematical reason for the error.

[1]
Question 47
HL • Paper 3
Hard
Calculator Permitted
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.

A

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

I.

Use the two data points to pp exactly.

[3]
II.

kk exactly.

[2]
B

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

I.

the mass of an animal with metabolic rate 300300 units.

[2]
II.

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

[2]
C

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

I.

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

[3]
II.

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

[2]
Question 48
HL • Paper 1
Hard
Non Calculator
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.

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

[3]
II.

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]
B

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

I.

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

[1]
II.

the exact values of aa and xx.

[3]
III.

log⁡xa\log_x a.

[2]

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Question 49
HL • Paper 1
Hard
Non Calculator
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, the equation may be written as 4u4−17u2+4=04u^4-17u^2+4=0.

[3]
II.

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

[3]
III.

the original equation.

[2]
B

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

[4]
Question 50
HL • Paper 1
Hard
Non Calculator
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]
II.

u2−14u+15=0u^2-14u+15=0.

[2]
B

the original equation, giving your answers in exact form.

[4]
C

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

[3]
Question 51
HL • Paper 1
Hard
Non Calculator
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. the equation can be written as u+1u=cu+\dfrac{1}{u}=c.

[3]
II.

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

[2]
B

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

[2]
C

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

[5]
Question 52
HL • Paper 1
Hard
Non Calculator
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]
II.

Q=0Q=0 for xx.

[2]
B

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

[3]
C

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

[3]
D

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

[2]

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Question 53
HL • Paper 2
Hard
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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}
A
I.

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

[2]
II.

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

[1]
B

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

[4]
C

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

[3]
Question 54
HL • Paper 2
Hard
Calculator Permitted
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. the equation can be written as u3−7u+5=0u^3-7u+5=0.

[3]
II.

why only positive roots of this cubic are relevant.

[1]
B

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

[2]
C

22x+5⋅2−x<72^{2x}+5\cdot2^{-x}<7.

[3]
Question 55
HL • Paper 2
Hard
Calculator Permitted
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.

ln⁡R=ln⁡A−BT\ln R=\ln A-\frac{B}{T}.

[1]
II.

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

[2]
B

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

[4]
C

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]
Question 56
HL • Paper 3
Hard
Calculator Permitted
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}, mm.

[2]
II.

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

[2]
B

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

I.

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

[3]
II.

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

[2]
C

A 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]
II.

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

[1]

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Question 57
HL • Paper 3
Hard
Calculator Permitted
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.

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.

the value of ln⁡A\ln A.

[1]
II.

AA, giving your answer in scientific notation.

[2]
III.

Given R=8.31R=8.31, EE.

[1]
B

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

I.

KK when T=350T=350 K.

[2]
II.

the temperature at which K=10.0K=10.0.

[2]
C

A 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. the gradient and intercept of this new line.

[3]
II.

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

[1]
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.

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

[2]
II.

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

[1]
Question 58
HL • Paper 3
Hard
Calculator Permitted
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.

A
I.

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

[2]
II.

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

[2]
B

Consider the sum of all nine probabilities.

I.

that ∑d=19P(d)=1\sum_{d=1}^{9}P(d)=1.

[3]
II.

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

[1]
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.

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

[3]
II.

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

[2]
Question 59
HL • Paper 3
Hard
Calculator Permitted
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.

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

[2]
II.

the transmission fraction of a filter with absorbance 1.301.30.

[2]
B

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

I.

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

[3]
II.

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

[1]
C

A 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.

the transmission fraction of one filter.

[1]
II.

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

[3]
III.

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

[2]
Question 60
HL • Paper 2
Hard
Calculator Permitted
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)
A
I.

the domain of ff.

[1]
II.

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

[1]
B

the inverse function is

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

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

[3]
D

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

[3]

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