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Exponent-Log Functions

Practice exam-style IB Math AA questions for Exponent-Log Functions, 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

Let f(x)=3⋅2x−5f(x)=3\cdot 2^x-5.

A

the equation of the horizontal asymptote of the graph of ff.

[1]
B

the yy-intercept of the graph of ff.

[1]
C

f(x)=19f(x)=19.

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

Let g(x)=ln⁡(x−2)+1g(x)=\ln(x-2)+1.

A

the domain of gg and the equation of its vertical asymptote.

[2]
B

the xx-intercept of the graph of gg.

[2]
C

the range of gg.

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

Let f(x)=5xf(x)=5^x.

A

Write f(x)f(x) in the form ekxe^{kx}.

[1]
B

f−1(x)f^{-1}(x), giving your answer using natural logarithms.

[2]
C

the exact value of f−1(255)f^{-1}(25\sqrt{5}).

[2]
Question 4
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

Consider the function f(x)=ln⁡(x−2)+3f(x)=\ln(x-2)+3.

A

the domain of ff and the equation of its vertical asymptote.

[2]
B

the xx-intercept of the graph of ff.

[2]
C

the value of xx for which f(x)=5f(x)=5.

[1]

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Question 5
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Consider the equation ln⁡(x+2)+ln⁡(x−1)=ln⁡10\ln(x+2)+\ln(x-1)=\ln 10.

A

the restriction on xx.

[1]
B

Write the equation as a quadratic equation in xx.

[2]
C

the equation.

[2]
Question 6
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A quantity QQ is modelled by Q(t)=AektQ(t)=Ae^{kt}, where tt is measured in hours. Initially Q=80Q=80, and after 33 hours Q=10Q=10.

A

the value of AA.

[1]
B

the exact value of kk.

[2]
C

the time at which Q=40Q=40.

[3]
Question 7
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let h(x)=log⁡2(x−1)+3h(x)=\log_2(x-1)+3.

A

the domain of hh and the equation of its vertical asymptote.

[2]
B

h(x)=5h(x)=5.

[2]
C

h−1(x)h^{-1}(x).

[2]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the equation e2x−5ex+6=0e^{2x}-5e^x+6=0.

A

By using the substitution u=exu=e^x, write the equation as a quadratic equation in uu.

[2]
B

the original equation for xx.

[3]

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Question 9
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A culture of bacteria is modelled by P(t)=AektP(t)=Ae^{kt}, where PP is the number of bacteria tt hours after observation begins. Initially there are 250250 bacteria. After 66 hours there are 410410 bacteria.

A

the value of AA.

[1]
B

the value of kk.

[2]
C

the number of bacteria after 1212 hours.

[1]
D

the time at which the model first predicts 10001000 bacteria.

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

A

the equation 3e2x−10ex+3=03e^{2x}-10e^x+3=0. Give your answers in exact form.

[5]
Question 11
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

An investment of 50005000 euros increases by 3.2%3.2\% each year. The value, VV euros, after tt years is modelled by V=5000(1.032)tV=5000(1.032)^t.

A

Write the model in the form V=5000ektV=5000e^{kt}, giving the value of kk.

[2]
B

the value of the investment after 88 years.

[2]
C

the time taken for the investment to double in value.

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

The graphs of y=exy=e^x and y=4−xy=4-x intersect at the point AA.

A

the equation that must be solved to the xx-coordinate of AA.

[1]
B

Use your graphic display calculator to the coordinates of AA.

[2]
C

why there is only one point of intersection.

[2]

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Question 13
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A phone battery is being charged. The percentage charge, BB, after tt hours is modelled by B(t)=100−80e−ktB(t)=100-80e^{-kt}, where kk is a positive constant. After 22 hours, the battery charge is 55%55\%.

A

the value of kk.

[2]
B

the time taken for the battery charge to reach 90%90\%.

[2]
C

According to the model, will the battery ever reach exactly 100%100\% charge? Give a reason for your answer.

[2]
Question 14
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the equation 2x+2−x=522^x+2^{-x}=\dfrac52.

A

Let u=2xu=2^x. Write the equation as a quadratic equation in uu.

[2]
B

the quadratic equation for uu.

[2]
C

all values of xx.

[1]
Question 15
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

log⁡x16+log⁡4x=3\log_x 16+\log_4 x=3, where x>0x>0 and x≠1x\neq 1.

A

Let y=log⁡4xy=\log_4 x. Express log⁡x16\log_x 16 in terms of yy.

[2]
B

Form a quadratic equation in yy.

[1]
C

for xx.

[2]
Question 16
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=aekx+bf(x)=ae^{kx}+b, where a>0a>0 and k>0k>0. The graph of ff has horizontal asymptote y=2y=2 and passes through the points (0,5)(0,5) and (ln⁡4,14)(\ln 4,14).

A

bb.

[1]
B

aa.

[1]
C

kk.

[3]

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Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Positive variables xx and yy are related by y=Cxny=Cx^n, where CC and nn are constants. It is known that y=6y=6 when x=1x=1, and y=48y=48 when x=4x=4.

A

CC.

[1]
B

the exact value of nn.

[3]
C

Write the relationship in the form ln⁡y=mln⁡x+c\ln y=m\ln x+c.

[1]
Question 18
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Let f(x)=axf(x)=a^x, where 0<a<10<a<1. The point (2,19)(2,\frac{1}{9}) lies on the graph of ff.

A

the value of aa.

[2]
B

an expression for f−1(x)f^{-1}(x).

[1]
C

The graph of ff meets the line y=xy=x at the point PP. the coordinates of PP.

[3]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A physical quantity yy is modelled by y=Cxny=Cx^n, where CC and nn are positive constants. Two experimental readings are y=10.4y=10.4 when x=3x=3, and y=83.1y=83.1 when x=12x=12.

A

By taking natural logarithms, a graph of ln⁡y\ln y against ln⁡x\ln x is a straight line.

[2]
B

Use the two readings to the value of nn.

[2]
C

the value of CC.

[1]
D

Use the model to yy when x=20x=20.

[2]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the inequality ln⁡(x+1)>12x−1\ln(x+1)>\frac{1}{2}x-1, where x>−1x>-1.

A

the two equations whose intersections can be used to the inequality graphically.

[1]
B

the xx-coordinates of the points of intersection.

[2]
C

the inequality.

[2]

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Question 21
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let p(x)=log⁡2(x+3)+log⁡2(7−x)p(x)=\log_2(x+3)+\log_2(7-x).

A

the domain of pp.

[1]
B

p(x)=4p(x)=4.

[3]
C

the maximum value of p(x)p(x).

[2]
Question 22
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let f(x)=axf(x)=a^x, where a>1a>1. The point (3,5)(3,5) lies on the graph of ff.

A

aa in exact form.

[1]
B

Write log⁡a20\log_a 20 in terms of natural logarithms, and its value.

[2]
C

all solutions to ax=x+2a^x=x+2.

[3]
Question 23
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

the equation log⁡3x+log⁡x3=52\log_3 x+\log_x 3=\frac{5}{2}, where x>0x>0 and x≠1x\neq1.

[5]
Question 24
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=5−4e−xf(x)=5-4e^{-x}, for x∈Rx\in\mathbb{R}.

A
I.

the yy-intercept of the graph of ff and the equation of its horizontal asymptote.

[3]
II.

the xx-intercept of the graph of ff.

[2]
B
I.

an expression for f−1(x)f^{-1}(x), stating its domain.

[3]
II.

f−1(x)=ln⁡2f^{-1}(x)=\ln 2.

[2]

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Question 25
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The temperature TT, in degrees Celsius, of a liquid after tt minutes is modelled by

T(t)=18+Ce−ktT(t)=18+Ce^{-kt}

where CC and kk are positive constants. Initially the temperature is 80∘C80^\circ\text{C}. After 44 minutes the temperature is 49∘C49^\circ\text{C}.

A
I.

(a)(i) the value of CC.

[2]
II.

(a)(ii) k=ln⁡24k=\dfrac{\ln 2}{4}.

[3]
B

Use C=62C=62 and k=ln⁡24k=\dfrac{\ln 2}{4} in this part. If you did not obtain this value of kk, use k=ln⁡24k=\dfrac{\ln 2}{4}.

I.

(b)(i) the exact time when T=1034T=\dfrac{103}{4}.

[3]
II.

(b)(ii) the horizontal asymptote of the graph of TT and its meaning in this context.

[3]
Question 26
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let h(x)=log⁡3(x−1)+log⁡3(10−x)h(x)=\log_3(x-1)+\log_3(10-x).

A
I.

the domain of hh.

[2]
II.

solving h(x)=2h(x)=2 leads to x2−11x+19=0x^2-11x+19=0.

[2]
B

If you did not obtain the quadratic in part (a), use x2−11x+19=0x^2-11x+19=0.

I.

h(x)=2h(x)=2.

[2]
II.

the maximum value of h(x)h(x).

[3]
Question 27
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let f(x)=log⁡axf(x)=\log_a x, where a>1a>1. The point (8,3)(8,3) lies on the graph of ff. A second function is defined by h(x)=log⁡2(x−3)+1h(x)=\log_2(x-3)+1.

A
I.

the value of aa.

[2]
II.

f−1(x)f^{-1}(x).

[2]
B
I.

the domain of hh and the equation of its vertical asymptote.

[2]
II.

the xx-intercept of the graph of hh.

[2]
III.

h−1(x)h^{-1}(x) and h−1(x)=11h^{-1}(x)=11.

[2]
Question 28
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=ln⁡(ex−1)f(x)=\ln(e^x-1), for x>0x>0.

A

the range of ff.

[1]
B

f−1(x)f^{-1}(x).

[3]
C

f−1(x)=ln⁡5f^{-1}(x)=\ln 5.

[2]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A cup of soup is left in a room where the temperature is 18∘C18^\circ\text{C}. The temperature of the soup, T∘CT^\circ\text{C}, tt minutes after it is left in the room is modelled by

T=18+Ae−kt,t≥0T=18+Ae^{-kt},\quad t\geq 0

where AA and kk are positive constants. Initially the temperature of the soup is 92∘C92^\circ\text{C}, and after 55 minutes its temperature is 61∘C61^\circ\text{C}.

A

the values of AA and kk.

[4]
B

Part (b): Using the model found in part (a):

I.

the temperature of the soup after 1212 minutes;

[2]
II.

the time taken for the temperature to reach 30∘C30^\circ\text{C}.

[2]
C

The average temperature of the soup during the first 2020 minutes is given by

Tˉ=120∫020T dt\bar T=\frac{1}{20}\int_0^{20}T\,\mathrm{d}t

this average temperature.

[2]
D

one limitation of using this model to the temperature of the soup after several hours.

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

The function ff is defined by

f(x)=aln⁡(x−1)+3,x>1f(x)=a\ln(x-1)+3,\quad x>1

where aa is a positive constant. The graph of ff passes through the point (6,7)(6,7).

A

:

I.

the exact value of aa;

[2]
II.

the equation of the vertical asymptote of the graph of ff.

[2]
B

the coordinates of the xx-intercept of the graph of ff.

[3]
C

The region RR is bounded by the graph of ff, the line y=3y=3, and the lines x=2x=2 and x=6x=6. the area of RR.

[2]
D

an expression for f−1(x)f^{-1}(x). its domain.

[2]
Question 31
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=7e−0.4x+2,x∈Rf(x)=7e^{-0.4x}+2,\quad x\in\mathbb{R}

Let PP be the point of intersection of the graph of ff with the line y=xy=x.

A

For the graph of ff:

I.

the equation of the horizontal asymptote;

[1]
II.

the yy-intercept;

[1]
III.

the range of ff.

[2]
B

f−1(x)f^{-1}(x) and its domain.

[3]
C

The graphs of y=f(x)y=f(x) and y=xy=x intersect at the point PP.

I.

the equation that must be solved to the xx-coordinate of PP.

[1]
II.

Use your graphic display calculator to the coordinates of PP.

[2]
Question 32
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The mass, MM grams, of a sample of yeast is modelled by

M=cert,t≥0M=ce^{rt},\quad t\geq0

where tt is measured in hours. The mass is 12.012.0 grams initially and 19.519.5 grams after 44 hours.

A

:

I.

the value of cc;

[1]
II.

the value of rr.

[3]
B

The model can also be written in the form M=12btM=12b^t. bb.

[2]
C

Using the model found in part (a):

I.

the mass after 1010 hours;

[1]
II.

the time taken for the mass to reach 5050 grams.

[2]
D

In a laboratory report, the student writes: "The yeast mass will continue to increase exponentially forever." on this statement in the context of the model.

[2]

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Question 33
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

A cup of liquid is removed from a heater and placed in a room. Its temperature, TT degrees Celsius, tt minutes later is modelled by

T(t)=18+Ae−kt,k>0T(t)=18+Ae^{-kt}, \quad k>0

The graph shows the cooling curve and its horizontal asymptote.

A
I.

Given that T(0)=82T(0)=82, AA.

[1]
II.

Given also that T(12)=50T(12)=50, k=ln⁡212k=\frac{\ln 2}{12}.

[4]
B
I.

the time at which the model predicts T=25T=25.

[3]
II.

the physical meaning of the horizontal asymptote in this model.

[1]
C

By taking logarithms, a graph of ln⁡(T−18)\ln(T-18) against tt is a straight line, and its gradient and vertical intercept.

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

The light level, LL lux, in a greenhouse after automatic shading is activated is modelled by

L(t)=C+Ae−kt,t≥0L(t)=C+Ae^{-kt}, \quad t\ge 0

where tt is measured in minutes. The graph shows LL approaching a limiting light level.

A
I.

The limiting light level is 4040 lux and L(0)=280L(0)=280. CC and AA.

[2]
II.

Given that L(15)=100L(15)=100, kk.

[3]
B
I.

an expression for tt in terms of LL, valid for 40<L≤28040<L\le 280.

[3]
II.

Use your expression to when the light level reaches 7070 lux.

[1]
C

A student claims that the model predicts the light level will be 4040 lux after a sufficiently long finite time. on the student's claim.

[3]
Question 35
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Positive variables xx and yy are related by y=Cxny=Cx^n, where CC and nn are constants. A graph of Y=ln⁡yY=\ln y against X=ln⁡xX=\ln x is a straight line passing through the points (ln⁡2,ln⁡12)(\ln 2,\ln 12) and (ln⁡8,ln⁡96)(\ln 8,\ln 96).

A
I.

the value of nn.

[3]
II.

the value of CC.

[2]
B

If you did not obtain the values in part (a), use C=32C=3\sqrt2 and n=32n=\dfrac32.

I.

Write the relationship between xx and yy explicitly.

[1]
II.

the value of xx when y=242y=24\sqrt2.

[3]
III.

a graph of ln⁡(y2)\ln(y^2) against ln⁡x\ln x has gradient 33 and its vertical intercept.

[3]
Question 36
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

For x∈Rx\in\mathbb{R}, define

F(x)=3x+9⋅3−xF(x)=3^x+9\cdot 3^{-x}
A
I.

By using the substitution u=3xu=3^x, solving F(x)=10F(x)=10 leads to u2−10u+9=0u^2-10u+9=0.

[3]
II.

F(x)=10F(x)=10.

[2]
B

If you did not part (a), use the fact that the solutions of F(x)=10F(x)=10 are x=0x=0 and x=2x=2.

I.

F(2−x)=F(x)F(2-x)=F(x) for all real xx.

[2]
II.

the value of xx at which FF is least, and this least value.

[2]

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Question 37
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let p>0p>0. Consider the equation

e2x−(p+4)ex+4p=0e^{2x}-(p+4)e^x+4p=0
A
AI.

Use the substitution u=exu=e^x to factorise the resulting quadratic equation in uu.

[3]
AII.

all real solutions for xx in terms of pp.

[3]
B

Use your result from part (a). If you did not obtain it, use the solutions x=ln⁡4x=\ln4 and x=ln⁡px=\ln p.

BI.

pp if one of the solutions is x=2ln⁡3x=2\ln3.

[2]
BII.

the values of pp for which the original equation has two distinct positive solutions for xx.

[2]
BIII.

the values of pp for which the original equation has exactly one positive solution for xx.

[2]
Question 38
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The function ff is defined by

f(x)=ln⁡(x−14−x),1<x<4f(x)=\ln\left(\frac{x-1}{4-x}\right),\qquad 1<x<4
A
I.

the equations of the vertical asymptotes of the graph of ff.

[2]
II.

the range of ff.

[1]
B
I.

an expression for f−1(x)f^{-1}(x).

[4]
II.

f−1(ln⁡2)f^{-1}(\ln2).

[1]
C
I.

f(5−x)=−f(x)f(5-x)=-f(x).

[3]
II.

the coordinates of the point about which the graph of ff has rotational symmetry.

[1]
Question 39
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For a>0a>0, define

pa(x)=ln⁡x+ln⁡(x+a),x>0p_a(x)=\ln x+\ln(x+a),\qquad x>0
A
I.

pa(x)=ln⁡(x2+ax)p_a(x)=\ln(x^2+ax).

[1]
II.

pa−1(x)p_a^{-1}(x) in terms of aa.

[4]
B

If you did not obtain the inverse in part (a), use pa−1(x)=−a+a2+4ex2p_a^{-1}(x)=\dfrac{-a+\sqrt{a^2+4e^x}}{2}.

I.

Given that pa−1(ln⁡24)=4p_a^{-1}(\ln24)=4, aa.

[3]
II.

For a=2a=2, p2(x)=2ln⁡3p_2(x)=2\ln3.

[3]
III.

why the other root of the quadratic in part (b)(ii) is rejected.

[1]
Question 40
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A medicine is eliminated from the bloodstream exponentially. The amount, CC mg, remaining tt hours after a dose of 8080 mg is given by

C=80e−kt,t≥0C=80e^{-kt},\quad t\geq0

The half-life of the medicine is 66 hours. A second dose of 8080 mg is taken 88 hours after the first dose.

A

the exact value of kk.

[3]
B

:

I.

the amount of medicine in the bloodstream immediately before the second dose;

[2]
II.

the amount of medicine in the bloodstream immediately after the second dose.

[2]
C

For t≥8t\geq8, the total amount of medicine in the bloodstream is

C=80e−kt+80e−k(t−8)C=80e^{-kt}+80e^{-k(t-8)}
[2]
D

If you did not obtain the expression in part (c), use C=80e−kt+80e−k(t−8)C=80e^{-kt}+80e^{-k(t-8)}. the time at which the total amount of medicine reaches 2020 mg after the second dose. , for which times the amount is below 2020 mg.

[3]

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Question 41
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The hydrogen ion concentration, HH mol dm−3\text{dm}^{-3}, in a solution is modelled by

H=Ae−kt,t≥0H=Ae^{-kt},\quad t\geq0

where tt is measured in minutes. The pH of the solution is defined by

pH=−log⁡10H\text{pH}=-\log_{10}H

At t=0t=0, the pH is 3.203.20. At t=12t=12, the pH is 4.104.10.

A

:

I.

the value of AA;

[2]
II.

the value of kk.

[2]
B

the pH can be written in the form

pH=3.20+ktln⁡10\text{pH}=3.20+\frac{kt}{\ln10}
[3]
C

Using the model:

I.

the time when the pH first reaches 5.005.00;

[2]
II.

the hydrogen ion concentration when t=20t=20.

[1]
D

the average hydrogen ion concentration during the first 2424 minutes.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A chemical indicator is diluted in equal stages. Let HnH_n be the concentration of hydrogen ions after nn dilution stages, where

Hn=H0rn,0<r<1H_n=H_0r^n, \quad 0<r<1

The pH value after nn stages is defined by

pn=−log⁡10(Hn)p_n=-\log_{10}(H_n)

nn

pH

0

3.40

1

3.65

2

3.90

3

4.15

4

4.40

5

4.65

6

4.90

7

5.15

8

5.40

9

5.65

12

6.40

A
I.

Write pnp_n in terms of H0H_0, rr and nn.

[3]
II.

why pnp_n is an arithmetic sequence.

[2]
B
I.

The initial pH is 3.403.40 and each stage increases the pH by 0.2500.250. rr.

[2]
II.

the least number of dilution stages required for the pH to exceed 6.006.00.

[2]
C

the converse result: if the pH values form an arithmetic sequence, then the concentrations HnH_n form a geometric sequence.

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

A narrow beam of light passes through layers of tinted glass. The intensity II after passing through thickness dd millimetres is modelled by

I(d)=I0e−kd,k>0I(d)=I_0e^{-kd},\quad k>0
A
AI.

Given I0=500I_0=500 and I(8)=180I(8)=180, kk.

[3]
AII.

Writing dd for the numerical value of the thickness in millimetres, write the model in the form I(d)=500adI(d)=500a^d, and aa.

[2]
B
BI.

the thickness required to reduce the intensity to 5%5\% of its initial value.

[3]
C

the additional thickness needed to reduce the intensity to rr times its current value, where 0<r<10<r<1, is independent of the starting intensity.

[4]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A laboratory colony is expected to increase by 60%60\% in one day if growth is compounded once daily. A researcher compares this with a model where the same nominal daily rate is compounded nn times per day:

Pn(t)=P0(1+0.60n)ntP_n(t)=P_0\left(1+\frac{0.60}{n}\right)^{nt}

where tt is measured in days.

A
I.

the once-daily model P1(t)P_1(t).

[1]
II.

The continuous model is P(t)=P0e0.60tP(t)=P_0e^{0.60t}. the percentage by which the continuous model exceeds the once-daily model after one day.

[3]
B
I.

For n=24n=24, the time taken for the colony to triple.

[3]
II.

the time taken for the colony to triple under the continuous model.

[1]
C

Pn(t)P_n(t) can be written in the form P0ekntP_0e^{k_nt}, and an expression for knk_n. why kn<0.60k_n<0.60 for finite nn.

[4]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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An experiment measures the mass mm grams of a volatile liquid remaining after tt minutes. The data are thought to follow an exponential model

m=Aekt,k<0m=Ae^{kt}, \quad k<0

A of ln⁡m\ln m against tt is provided.

A
I.

the model can be linearised by plotting ln⁡m\ln m against tt.

[2]
II.

what the gradient and vertical intercept of this line represent.

[2]
B
I.

Using linear regression on the transformed data, the line of best fit is

ln⁡m=3.91−0.200t\ln m=3.91-0.200t

the exponential model for mm.

[2]
II.

Use the model to the time when the mass first falls below 1010 grams.

[3]
C

The measured mass at t=4t=4 minutes is 22.522.5 grams. the residual, defined as measured mass minus modelled mass, and its sign.

[4]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The energy EE released by a seismic event of magnitude MM is modelled by

E=E0×101.5ME=E_0\times 10^{1.5M}

where E0E_0 is a positive constant. This question investigates how logarithms are used to events.

A
I.

Let d>0d>0. If the lower-magnitude event has magnitude MM and the higher-magnitude event has magnitude M+dM+d, the ratio of the energy of the higher-magnitude event to that of the lower-magnitude event is 101.5d10^{1.5d}.

[3]
II.

the energy ratio for a magnitude difference of 0.400.40.

[1]
B
I.

the magnitude of a single event that releases the same energy as 2020 events, each of magnitude 4.64.6.

[3]
II.

why the answer is not 20×4.620\times4.6.

[1]
C

Generalise part (b): the magnitude MnM_n of one event equivalent in energy to nn events, each of magnitude MM, where n∈Z+n\in\mathbb{Z}^+.

[4]
Question 47
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For x>0x>0, x≠1x\ne1, consider the equation

log⁡x4+log⁡2x=k\log_x 4+\log_2 x=k

where kk is a real constant.

A
I.

Let y=log⁡2xy=\log_2 x. Express log⁡x4\log_x 4 in terms of yy.

[3]
II.

the equation can be written as y2−ky+2=0y^2-ky+2=0.

[2]
B

If you did not obtain the quadratic in part (a), use y2−ky+2=0y^2-ky+2=0.

I.

the original equation when k=3k=3.

[2]
II.

the original equation when k=−3k=-3.

[2]
III.

the values of kk for which the original equation has real solutions.

[4]
Question 48
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Consider positive real numbers xx and yy satisfying the system

2x4y=32,log⁡2x+log⁡2y=12^x4^y=32,\qquad \log_2 x+\log_2 y=1
A
I.

the system is equivalent to x+2y=5x+2y=5 and xy=2xy=2.

[3]
II.

the system.

[2]
B

For parts (b)(i) and (b)(ii), consider the generalized system

2x4y=2n,log⁡2x+log⁡2y=12^x4^y=2^n,\qquad \log_2 x+\log_2 y=1

where nn is a real constant.

I.

yy satisfies 2y2−ny+2=02y^2-ny+2=0.

[2]
II.

the values of nn for which the system has two distinct positive solutions.

[4]

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Question 49
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For 0<x<80<x<8, define

q(x)=log⁡2x+log⁡2(8−x)q(x)=\log_2 x+\log_2(8-x)
A
I.

q(x)=log⁡2(16−(x−4)2)q(x)=\log_2\left(16-(x-4)^2\right).

[3]
II.

the maximum value of q(x)q(x) and the value of xx at which it occurs.

[2]
B

If you did not obtain the result in part (a), use q(x)=log⁡2(16−(x−4)2)q(x)=\log_2\left(16-(x-4)^2\right).

I.

q(x)=3q(x)=3.

[3]
II.

q(x)=4q(x)=4.

[2]
C

Let rr be a real constant. the number of solutions of q(x)=rq(x)=r in the interval 0<x<80<x<8.

I.

For a real constant rr, the number of solutions of q(x)=rq(x)=r in 0<x<80<x<8, and your answer.

[4]
Question 50
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let

f(x)=ln⁡(x+3),x>−3f(x)=\ln(x+3),\quad x>-3

and

g(x)=2e−0.5xg(x)=2e^{-0.5x}

The graphs of y=f(x)y=f(x) and y=g(x)y=g(x) intersect at the point AA.

A

the coordinates of AA.

[3]
B

Let h(x)=f(x)−g(x)=ln⁡(x+3)−2e−0.5xh(x)=f(x)-g(x)=\ln(x+3)-2e^{-0.5x}.

I.

h′(x)h'(x).

[2]
II.

that the graphs of ff and gg intersect exactly once.

[2]
C

If you did not obtain the xx-coordinate of AA, use x=0.806x=0.806. the area enclosed by the graphs of ff and gg and the yy-axis.

[3]
D

the value of xx for which the vertical distance between the two graphs is greatest on the interval 0≤x≤0.8060\leq x\leq0.806.

[2]
Question 51
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

An experiment suggests that two positive variables xx and yy are related by a power model

y=Cxny=Cx^n

where CC and nn are positive constants. The following table gives four measurements.

x

y

2

10.7

3

18.5

5

36.9

8

69.6

A

By taking natural logarithms, a graph of ln⁡y\ln y against ln⁡x\ln x should be a straight line. the gradient and vertical intercept in terms of CC and nn.

[4]
B

A least-squares linear regression of ln⁡y\ln y on ln⁡x\ln x using the four measurements gives

ln⁡y≈1.434+1.351ln⁡x\ln y\approx1.434+1.351\ln x

Use this regression equation to :

I.

the value of nn;

[1]
II.

the value of CC;

[2]
III.

the model for yy in terms of xx.

[1]
C

Use the model to the value of xx when y=80y=80.

[2]
D

A student claims that doubling xx will double yy. Use the model to on this claim.

[3]
Question 52
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function ff is defined by

f(x)=ln⁡(3+e−x),x∈Rf(x)=\ln(3+e^{-x}),\quad x\in\mathbb{R}
A

For the graph of ff:

I.

the equation of the horizontal asymptote as x→∞x\to\infty;

[1]
II.

the range of ff.

[3]
B

an expression for f−1(x)f^{-1}(x) and its domain.

[3]
C

The graph of ff intersects the line y=xy=x at the point PP.

I.

the xx-coordinate of PP satisfies

e2x−3ex−1=0e^{2x}-3e^x-1=0
[2]
II.

the coordinates of PP.

[1]
D

the area between the graph of ff and its horizontal asymptote from x=0x=0 to x=4x=4.

[2]

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Question 53
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A patient receives repeated doses of a medicine. Immediately after each dose, 5050 mg is added to the amount already in the bloodstream. Between doses, the amount decreases exponentially with decay constant 0.180.18 per hour. Doses are taken every 66 hours.

Let CnC_n be the amount of medicine in the bloodstream, in mg, immediately after the nnth dose.

Assume that there is no medicine in the bloodstream immediately before the first dose.

A

:

I.

the amount immediately before the second dose;

[1]
II.

the value of C2C_2.

[2]
B

Cn+1=e−1.08Cn+50C_{n+1}=e^{-1.08}C_n+50
[3]
C

If you did not obtain the recurrence in part (b), use Cn+1=e−1.08Cn+50C_{n+1}=e^{-1.08}C_n+50.

I.

C5C_5.

[2]
II.

the limiting value of CnC_n as n→∞n\to\infty.

[2]
D

In the long term, the amount of medicine in the bloodstream immediately before a dose is taken.

[2]
E

A safety guideline states that the amount immediately after a dose should remain below 9090 mg. According to this model, whether the guideline will be met in the long term.

[1]
Question 54
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

This question investigates intersections of the exponential graph y=axy=a^x with the line y=xy=x, where a>1a>1. The diagram shows the line y=xy=x and several graphs of y=axy=a^x for different values of aa.

A
I.

Suppose that y=axy=a^x is tangent to y=xy=x at x=tx=t. t=et=e.

[4]
II.

the value of aa for which the graph is tangent to the line.

[2]
B
I.

For a=1.2a=1.2, use your GDC to the two intersection points of y=1.2xy=1.2^x and y=xy=x.

[2]
C

the number of intersections of y=axy=a^x and y=xy=x for all a>1a>1. If you did not obtain a=e1/ea=e^{1/e} in part (a), use this value.

[4]
Question 55
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

This question investigates the intersections of y=exy=e^x and the family of straight lines y=mxy=mx, where m>0m>0.

A
I.

if y=mxy=mx is tangent to y=exy=e^x, then the point of tangency has xx-coordinate 11.

[3]
II.

the corresponding value of mm.

[2]
B
I.

For m=4m=4, use your GDC to the two solutions of ex=4xe^x=4x.

[2]
II.

why there are no negative solutions for any m>0m>0.

[1]
C

the number of solutions of ex=mxe^x=mx for all m>0m>0. If you did not obtain m=em=e in part (a), use this value.

[4]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a real parameter pp, define

fp(x)=p+ln⁡x,x>0f_p(x)=p+ \ln x, \quad x>0

This question investigates fixed points of fpf_p, that is, solutions of fp(x)=xf_p(x)=x.

A
I.

For p=2p=2, use your GDC to the fixed points of fpf_p.

[2]
II.

For p=1p=1, x=1x=1 is a fixed point.

[2]
B
I.

fixed points of fpf_p are solutions of x−ln⁡x=px- \ln x=p.

[1]
II.

the minimum value of g(x)=x−ln⁡xg(x)=x- \ln x for x>0x>0.

[3]
C

the number of fixed points of fpf_p for all real values of pp. If you did not obtain the minimum value 11, use this value.

[4]

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Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let a>0a>0, a≠1a\ne1, and define

Fa(x)=log⁡ax+log⁡xaF_a(x)=\log_a x+ \log_x a

where x>0x>0 and x≠1x\ne1. This question investigates the possible values of Fa(x)F_a(x).

A
I.

Use the change-of-base formula to if u=log⁡axu=\log_a x, then Fa(x)=u+1uF_a(x)=u+\frac{1}{u}.

[3]
II.

the possible values of uu.

[2]
B
I.

Fa(x)=5F_a(x)=5 in terms of aa.

[3]
C

all possible values of Fa(x)F_a(x).

[4]
Question 58
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For a positive constant kk, consider the equation

ekx=xe^{kx}=x

For part (a), let k=0.2k=0.2. For parts (b) to (d), kk is an arbitrary positive constant.

A

For k=0.2k=0.2:

I.

the equation to be solved;

[1]
II.

use your graphic display calculator to all solutions.

[3]
B

Let F(x)=ekx−xF(x)=e^{kx}-x.

I.

the stationary point of FF occurs at

x=−ln⁡kkx=-\frac{\ln k}{k}
[3]
II.

the value of FF at this stationary point is

1+ln⁡kk\frac{1+\ln k}{k}
[1]
C

Use part (b) to the values of kk for which the equation ekx=xe^{kx}=x has two distinct real solutions.

[4]
D

what happens when k=e−1k=e^{-1}.

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

For x>0x>0, define

p(x)=ln⁡x+ln⁡(10−x),0<x<10p(x)=\ln x+\ln(10-x),\quad 0<x<10

The graph of pp is used to model a quantity that depends on the product of two positive factors whose sum is 1010.

A

the domain of pp and write p(x)p(x) as a single logarithm.

[3]
B

the maximum value of p(x)p(x).

[4]
C

Let mm be a real constant. the values of mm for which the equation p(x)=mp(x)=m has exactly two solutions.

[3]
D

For m=3m=3, p(x)=3p(x)=3.

[2]
E

If you did not obtain the solutions in part (d), use x=2.78x=2.78 and x=7.22x=7.22. the area under the graph of pp between these two solutions.

[2]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

After a short injection, the concentration CC of a tracer in a bloodstream is modelled by

C(t)=M(e−0.18t−e−0.55t),t≥0C(t)=M\left(e^{-0.18t}-e^{-0.55t}\right), \quad t\ge0

where tt is measured in minutes and M>0M>0. The graph shows a single peak in concentration.

A
I.

Given that C(2)=3.00C(2)=3.00, MM.

[3]
II.

why C(0)=0C(0)=0.

[1]
B
I.

the time of maximum concentration satisfies

e0.37t=0.550.18e^{0.37t}=\frac{0.55}{0.18}
[4]
II.

the time and value of the maximum concentration. If you did not obtain the equation in part (b)(i), use it here.

[3]
C

Use your GDC to the time interval during which C(t)>1.00C(t)>1.00.

[4]

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Modulus & Inequalities