Level 1 — RecognitionExponentials & Logarithms

Exponentials & Logarithms

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Choose the single best answer.

Q1. The horizontal asymptote of the graph of $y = 2^x$ is:

(A) $x = 0$ (B) $y = 0$ (C) $y = 1$ (D) $y = 2$

Q2. The value of $\log_3 81$ is:

(A) $3$ (B) $4$ (C) $27$ (D) $9$

Q3. Simplify $\dfrac{a^5}{a^2}$ (for $a \neq 0$ ):

(A) $a^{10}$ (B) $a^{7}$ (C) $a^{3}$ (D) $a^{2.5}$

Q4. The number $e$ is defined as:

(A) $\lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^n$ (B) $\lim_{n\to 0}\left(1+\tfrac{1}{n}\right)^n$ (C) $\lim_{n\to\infty}\left(1-\tfrac{1}{n}\right)^n$ (D) exactly $2.7$

Q5. Which of these equals $\ln x$ ?

(A) $\log_{10} x$ (B) $\log_e x$ (C) $\log_2 x$ (D) $\log_x e$

Q6. $\log_a M + \log_a N$ equals:

(A) $\log_a(M+N)$ (B) $\log_a(MN)$ (C) $\log_a(M/N)$ (D) $(\log_a M)(\log_a N)$

Q7. The change of base formula gives $\log_a b =$ :

(A) $\dfrac{\log b}{\log a}$ (B) $\dfrac{\log a}{\log b}$ (C) $\log b - \log a$ (D) $\log b \cdot \log a$

Q8. If a quantity decays with half-life $T$ , after time $2T$ the fraction remaining is:

(A) $\tfrac{1}{2}$ (B) $\tfrac{1}{3}$ (C) $\tfrac{1}{4}$ (D) $0$

Q9. The solution of $\log_2 x = 5$ is:

(A) $10$ (B) $25$ (C) $32$ (D) $7$

Q10. The domain of $y = \ln x$ is:

(A) all real $x$ (B) $x > 0$ (C) $x \geq 0$ (D) $x \neq 0$

Section B — Matching (1 mark each; 5 marks total)

Q11. Match each expression/quantity (left) to its correct value or description (right).

#	Item		Option
a	$\log_{10} 1000$	i	$0$
b	$\log_5 1$	ii	measures acidity
c	pH	iii	$3$
d	Richter scale	iv	measures earthquake magnitude
e	$e^0$	v	$1$

(Write pairs, e.g. a–iii)

Section C — True/False WITH Justification (2 marks each: 1 mark T/F, 1 mark reason)

Q12. $\log_a(M^k) = k\log_a M$ for valid $M, a$ .

Q13. The graph of $y = a^x$ (with $a>1$ ) passes through the point $(0,1)$ .

Q14. $\log(A) - \log(B) = \log(A - B)$ .

Q15. The graphs of $y = e^x$ and $y = \ln x$ are reflections of each other in the line $y = x$ .

Q16. For $0 < a < 1$ , the function $y = a^x$ is increasing.

Q17. Doubling time is the time for an exponentially growing quantity to increase by a factor of 2.

Q18. $\log_{10} 0 = 1$ .

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (B) $y=0$ — As $x\to-\infty$ , $2^x\to0$ ; asymptote is the horizontal line $y=0$ . (1)

Q2. (B) $4$ — $3^4 = 81$ , so $\log_3 81 = 4$ . (1)

Q3. (C) $a^3$ — Quotient law: $a^5/a^2 = a^{5-2}=a^3$ . (1)

Q4. (A) — $e=\lim_{n\to\infty}\left(1+\tfrac1n\right)^n \approx 2.718$ . Option (C) gives $1/e$ ; (D) is only approximate. (1)

Q5. (B) $\log_e x$ — Natural log is log to base $e$ . (1)

Q6. (B) $\log_a(MN)$ — Product law of logarithms. (1)

Q7. (A) $\dfrac{\log b}{\log a}$ — Change of base: $\log_a b = \dfrac{\log_c b}{\log_c a}$ . (1)

Q8. (C) $\tfrac14$ — After one half-life $\tfrac12$ , after two $\left(\tfrac12\right)^2=\tfrac14$ . (1)

Q9. (C) $32$ — $\log_2 x=5 \Rightarrow x=2^5=32$ . (1)

Q10. (B) $x>0$ — Logarithm defined only for positive arguments. (1)

Section B

Q11. a–iii ( $\log_{10}1000=3$ ); b–i ( $\log_5 1=0$ ); c–ii (pH measures acidity); d–iv (Richter measures earthquakes); e–v ( $e^0=1$ ). (1 each, 5 total)

Section C (1 mark T/F + 1 mark justification)

Q12. TRUE — Power law of logarithms; proof: let $x=\log_a M$ so $M=a^x$ , then $M^k=a^{kx}$ , giving $\log_a(M^k)=kx=k\log_a M$ . (2)

Q13. TRUE — $a^0=1$ for any $a>0$ , so every exponential $y=a^x$ passes through $(0,1)$ . (2)

Q14. FALSE — The quotient law gives $\log A-\log B=\log(A/B)$ , not $\log(A-B)$ . (2)

Q15. TRUE — $\ln x$ is the inverse of $e^x$ ; inverse function graphs are reflections in $y=x$ . (2)

Q16. FALSE — For $0<a<1$ , $a^x$ is decreasing (e.g. $(1/2)^x$ ). (2)

Q17. TRUE — Doubling time is defined as the time for the quantity to multiply by 2. (2)

Q18. FALSE — $\log_{10}0$ is undefined (no power of 10 equals 0); as $x\to0^+$ , $\log x\to-\infty$ . (2)

[
  {"claim":"log_3 81 = 4","code":"result = (log(81,3)==4)"},
  {"claim":"a^5/a^2 = a^3","code":"a=symbols('a',positive=True); result = simplify(a**5/a**2 - a**3)==0"},
  {"claim":"log_2 32 = 5","code":"result = (log(32,2)==5)"},
  {"claim":"half-life two periods leaves 1/4","code":"result = (Rational(1,2)**2 == Rational(1,4))"},
  {"claim":"change of base log_a b = ln b / ln a","code":"a,b=symbols('a b',positive=True); result = simplify(log(b,a) - log(b)/log(a))==0"}
]