Level 2 — RecallExponentials & Logarithms

Exponentials & Logarithms

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level: 2 (Recall — definitions, standard problems, short derivations) Time limit: 30 minutes Total marks: 40

Q1. State the value of each. (3 marks) (a) $\log_2 32$ (b) $\log_{10} 0.001$ (c) $\ln e^5$

Q2. Simplify using the laws of exponents, leaving your answer with positive indices. (3 marks) $\frac{x^{5}\,x^{-2}}{x^{3}} \cdot \left(x^{\frac12}\right)^{4}$

Q3. Write as a single logarithm: (3 marks) $2\log_{10} 3 + \log_{10} 5 - \log_{10} 9$

Q4. Solve for $x$ , giving your answer to 3 significant figures. (4 marks) $5^{x} = 20$

Q5. Solve the logarithmic equation. (4 marks) $\log_{2}(x+3) + \log_{2}(x-3) = 4$

Q6. Prove the product law of logarithms: for $M,N>0$ , (4 marks) $\log_a(MN) = \log_a M + \log_a N.$

Q7. Prove the change-of-base formula $\displaystyle \log_a b = \frac{\log_c b}{\log_c a}$ . (4 marks)

Q8. A radioactive substance has a half-life of 8 days. (5 marks) (a) Write a model $A(t) = A_0 (\tfrac12)^{t/8}$ and state what each symbol means. (2) (b) What fraction remains after 24 days? (1) (c) Find the time for the substance to decay to 10% of its initial amount (3 s.f.). (2)

Q9. Sketch and describe the graph of $y = e^{x}$ . State its horizontal asymptote and the coordinates where it crosses the $y$ -axis. (4 marks)

Q10. The pH of a solution is defined by $\text{pH} = -\log_{10}[\text{H}^+]$ , where $[\text{H}^+]$ is the hydrogen-ion concentration in mol L⁻¹. (6 marks) (a) Find the pH when $[\text{H}^+] = 1\times10^{-4}$ . (2) (b) Find $[\text{H}^+]$ when pH $= 8.5$ (2 s.f.). (2) (c) By how many times does $[\text{H}^+]$ change when pH decreases by 2? (2)

Answer keyMark scheme & solutions

Q1. (3 marks) (a) $2^5 = 32 \Rightarrow \log_2 32 = 5$ (1) (b) $10^{-3}=0.001 \Rightarrow \log_{10}0.001 = -3$ (1) (c) $\ln e^5 = 5\ln e = 5$ (1)

Q2. (3 marks) Numerator exponents: $5-2 = 3$ , so $x^3$ . Divide by $x^3$ : $x^{3-3}=x^0=1$ . (1) $(x^{1/2})^4 = x^2$ . (1) Product: $1\cdot x^2 = x^2$ . (1)

Q3. (3 marks) $2\log_{10}3 = \log_{10}9$ (power rule). (1) $\log_{10}9 + \log_{10}5 = \log_{10}45$ (product). (1) $\log_{10}45 - \log_{10}9 = \log_{10}5$ (quotient). (1)

Q4. (4 marks) Take logs: $x\ln 5 = \ln 20$ . (2) $x = \dfrac{\ln 20}{\ln 5} = \dfrac{2.9957}{1.6094} = 1.8614$ . (1) $x \approx 1.86$ (3 s.f.). (1)

Q5. (4 marks) Combine: $\log_2[(x+3)(x-3)] = 4$ . (1) $(x+3)(x-3) = 2^4 = 16 \Rightarrow x^2 - 9 = 16$ . (1) $x^2 = 25 \Rightarrow x = \pm 5$ . (1) Domain requires $x>3$ , so reject $x=-5$ ; answer $x = 5$ . (1)

Q6. (4 marks) Let $x = \log_a M,\ y = \log_a N$ , so $a^x = M,\ a^y = N$ . (1) Then $MN = a^x a^y = a^{x+y}$ (law of exponents). (1) Taking $\log_a$ : $\log_a(MN) = x+y$ . (1) $= \log_a M + \log_a N$ . ∎ (1)

Q7. (4 marks) Let $x = \log_a b$ , so $a^x = b$ . (1) Take $\log_c$ of both sides: $\log_c(a^x) = \log_c b$ . (1) $x\log_c a = \log_c b$ (power rule). (1) $x = \dfrac{\log_c b}{\log_c a}$ , i.e. $\log_a b = \dfrac{\log_c b}{\log_c a}$ . ∎ (1)

Q8. (5 marks) (a) $A(t)=A_0(\tfrac12)^{t/8}$ : $A_0$ = initial amount, $A(t)$ = amount at time $t$ (days), 8 = half-life. (2) (b) $t=24=3$ half-lives $\Rightarrow (\tfrac12)^3 = \tfrac18$ remains. (1) (c) $(\tfrac12)^{t/8}=0.1 \Rightarrow \tfrac{t}{8}\ln\tfrac12 = \ln 0.1$ $t = 8\cdot\dfrac{\ln 0.1}{\ln 0.5} = 8\times 3.3219 = 26.575 \approx 26.6$ days. (2)

Q9. (4 marks) Increasing curve for all $x$ , always positive. (1) Horizontal asymptote $y=0$ as $x\to-\infty$ . (1) Crosses $y$ -axis at $(0,1)$ . (1) Rises steeply for $x>0$ ; concave up throughout / correct shape sketch. (1)

Q10. (6 marks) (a) $\text{pH} = -\log_{10}(10^{-4}) = 4$ . (2) (b) $[\text{H}^+] = 10^{-8.5} = 3.16\times10^{-9} \approx 3.2\times10^{-9}$ mol L⁻¹. (2) (c) pH decrease of 2 means $-\log[\text{H}^+]$ drops by 2, so $[\text{H}^+]$ multiplied by $10^{2}=100$ (increases 100 times). (2)

[
  {"claim":"Q4: log20/log5 ≈ 1.86","code":"v=ln(20)/ln(5); result = abs(float(v)-1.86)<0.005"},
  {"claim":"Q5: x=5 satisfies equation","code":"x=5; result = simplify(log(x+3,2)+log(x-3,2)-4)==0"},
  {"claim":"Q8c: decay time ≈ 26.6 days","code":"t=8*ln(Rational(1,10))/ln(Rational(1,2)); result = abs(float(t)-26.6)<0.1"},
  {"claim":"Q10b: 10**-8.5 ≈ 3.2e-9","code":"v=Rational(10)**Rational(-17,2); result = abs(float(v)-3.2e-9)<0.1e-9"}
]