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A-Level Pure 2 · Edexcel IAL WMA12 & CAIE 9709 · Topic pack

Topic pack · Exponentials & logarithms แพ็กฝึกเฉพาะหัวข้อ · เอกซ์โพเนนเชียลและลอการิทึม

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Log laws, equations & models, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

Write $2\log_2 6-\log_2 9$ as a single logarithm in the form $\log_2 k$, where $k$ is an integer, and hence show that $2\log_2 6-\log_2 9=2$.

Mark schemeshow ▾
M1 power law: $2\log_2 6=\log_2 36$  M1 quotient law: $\log_2 36-\log_2 9=\log_2\tfrac{36}{9}=\log_2 4$  A1 $\log_2 4=2$ ∎.
Q2. 4 marks

Solve, giving each answer to 3 significant figures: (a) $3^x=20$   (b) $5^{2x-1}=8$.

Mark schemeshow ▾
(a) M1 take logs: $x=\dfrac{\log 20}{\log 3}$  A1 $x=2.73$ (exact $2.72683\ldots$).
(b) M1 $2x-1=\dfrac{\log 8}{\log 5}=1.29203\ldots$  A1 $x=\dfrac{1+1.29203}{2}=1.15$ (exact $1.14601\ldots$).
Q3. 4 marks

Solve $\log_2 x+\log_2(x-6)=4$.

Mark schemeshow ▾
M1 product law: $\log_2\big(x(x-6)\big)=4$, so $x(x-6)=2^4=16$  M1 $x^2-6x-16=0 \Rightarrow (x-8)(x+2)=0$  A1 $x=8$  B1 reject $x=-2$: both $\log_2 x$ and $\log_2(x-6)$ need positive arguments, so $x>6$.
Q4. 5 marks

Solve $4^x-5(2^x)+4=0$.

Mark schemeshow ▾
M1 let $y=2^x$, so $4^x=(2^x)^2=y^2$  A1 $y^2-5y+4=0$  M1 $(y-1)(y-4)=0$  A1 $y=1$ or $y=4$  A1 $2^x=1\Rightarrow x=0$; $2^x=4\Rightarrow x=2$.
Q5. 5 marks

Solve $2^{x+1}=3^{2x-1}$, giving your answer to 3 significant figures.

Mark schemeshow ▾
M1 take logs of both sides  A1 $(x+1)\log 2=(2x-1)\log 3$  M1 collect $x$ terms: $x(2\log 3-\log 2)=\log 3+\log 2$  A1 $x=\dfrac{\log 6}{\log 4.5}$  A1 $x=1.19$ (exact $1.19127\ldots$).
Q6. 6 marks

The number of bacteria in a culture is modelled by $N=400e^{0.15t}$, where $t$ is the time in hours after observation begins. (a) State the initial number of bacteria. (b) Find the number of bacteria after 5 hours, to the nearest whole number. (c) Find, to 3 significant figures, the time at which $N=2000$.

Mark schemeshow ▾
(a) B1 $t=0 \Rightarrow N=400$.
(b) M1 $N=400e^{0.75}$  A1 $=846.80\ldots \approx 847$.
(c) M1 $e^{0.15t}=5$  M1 $t=\dfrac{\ln 5}{0.15}$  A1 $t=10.7$ hours (exact $10.7296\ldots$).
Q7. 6 marks

(a) Solve $2^{3x-1}=10$, giving your answer to 3 significant figures. (b) Solve $\log_2 y=3+\log_2(y-7)$.

Mark schemeshow ▾
(a) M1 $3x-1=\dfrac{\log 10}{\log 2}=3.32193\ldots$  A1 $x=\dfrac{1+3.32193}{3}$  A1 $x=1.44$ (exact $1.44064\ldots$).
(b) M1 $\log_2\dfrac{y}{y-7}=3 \Rightarrow y=8(y-7)$  A1 $7y=56$, $y=8$  B1 check: $y-7=1>0$, so $y=8$ is valid.
Q8. 7 marks

The mass $M$ grams of a radioactive substance is modelled by $M=80e^{-kt}$, where $t$ is the time in years and $k$ is a positive constant. When $t=5$, the mass is 60 grams. (a) Find $k$ to 3 significant figures. (b) Find the mass when $t=12$, to 3 significant figures. (c) Find, to 3 significant figures, the time at which the mass reaches 20 grams.

Mark schemeshow ▾
(a) M1 $60=80e^{-5k} \Rightarrow e^{-5k}=0.75$  M1 $-5k=\ln 0.75$, so $k=\tfrac{1}{5}\ln\tfrac{4}{3}$  A1 $k=0.0575$ (exact $0.0575364\ldots$).
(b) M1 $M=80e^{-12k}=80\times 0.75^{12/5}$  A1 $M=40.1$ g (exact $40.1086\ldots$).
(c) M1 $e^{-kt}=\tfrac{1}{4} \Rightarrow t=\dfrac{\ln 4}{k}$  A1 $t=24.1$ years (exact $24.0942\ldots$).
Q9. 7 marks

Solve $3^{2x+1}-28(3^x)+9=0$.

Mark schemeshow ▾
M1 $3^{2x+1}=3\,(3^x)^2$  M1 let $y=3^x$: $3y^2-28y+9=0$  A1 quadratic correct  M1 $(3y-1)(y-9)=0$  A1 $y=\tfrac{1}{3}$ or $y=9$  A1 $3^x=\tfrac13 \Rightarrow x=-1$  A1 $3^x=9 \Rightarrow x=2$.
Q10. 8 marks

The values, in pounds, of two cars are modelled by $V_A=24000e^{-0.15t}$ and $V_B=16000e^{-0.08t}$, where $t$ is the age in years. (a) State the value of each car when new. (b) Find the value of car A after 4 years, to 3 significant figures. (c) Show that when the two cars have equal value, $e^{0.07t}=1.5$. (d) Hence find, to 3 significant figures, the age at which the two cars have equal value.

Mark schemeshow ▾
(a) B1 $t=0$: car A £24000, car B £16000.
(b) M1 $V_A=24000e^{-0.6}$  A1 £13200 (exact $13171.5\ldots$).
(c) M1 $24000e^{-0.15t}=16000e^{-0.08t}$  M1 divide: $\dfrac{24000}{16000}=e^{0.15t-0.08t}$  A1 $e^{0.07t}=1.5$ ∎.
(d) M1 $0.07t=\ln 1.5$  A1 $t=5.79$ years (exact $5.79236\ldots$).