LAWS OF EXPONENTS & INTRODUCTION TO LOGARITHMS
This question bank contains 11 questions covering laws of exponents and introduction to logarithms, distributed across different paper types according to IB AAHL curriculum standards.
📌 Multiple Choice Questions (2 Questions)
MCQ 1. Simplify: \(\frac{3^7 \times 3^2}{3^5}\)
A) \(3^2\) B) \(3^4\) C) \(3^9\) D) \(3^{14}\)
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Solution:
Step 1: Apply product law in numerator: \(3^7 \times 3^2 = 3^{7+2} = 3^9\)
Step 2: Apply quotient law: \(\frac{3^9}{3^5} = 3^{9-5} = 3^4\)
✅ Answer: B) \(3^4\)
MCQ 2. What is the value of \(\log_5 125\)?
A) 2 B) 3 C) 5 D) 25
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Solution:
By definition: \(\log_5 125 = x\) means \(5^x = 125\)
Since \(125 = 5^3\), we have \(5^x = 5^3\)
Therefore \(x = 3\)
✅ Answer: B) 3
📌 Paper 1 Questions (No Calculator) – 4 Questions
Paper 1 – Q1. Simplify: \((2x^3)^4 \times (3x^2)^2\)
[4 marks]
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Solution:
Step 1: Apply power law to each term
\((2x^3)^4 = 2^4 \times (x^3)^4 = 16x^{12}\)
\((3x^2)^2 = 3^2 \times (x^2)^2 = 9x^4\)
Step 2: Multiply the results
\(16x^{12} \times 9x^4 = 16 \times 9 \times x^{12} \times x^4\)
\(= 144x^{16}\)
✅ Answer: \(144x^{16}\)
Paper 1 – Q2. Solve for x: \(2^{x+1} = 16\)
[3 marks]
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Solution:
Step 1: Express 16 as a power of 2
\(16 = 2^4\)
Step 2: Rewrite equation
\(2^{x+1} = 2^4\)
Step 3: Equate exponents
\(x + 1 = 4\)
\(x = 3\)
✅ Answer: x = 3
Paper 1 – Q3. Find the exact value of \(\log_3 \frac{1}{27}\)
[3 marks]
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Solution:
Step 1: Express \(\frac{1}{27}\) as a power of 3
Since \(27 = 3^3\), we have \(\frac{1}{27} = \frac{1}{3^3} = 3^{-3}\)
Step 2: Apply logarithm definition
\(\log_3 3^{-3} = -3\)
✅ Answer: -3
Paper 1 – Q4. Solve for x: \(\log_2(x + 5) = 3\)
[4 marks]
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Solution:
Step 1: Convert to exponential form
\(\log_2(x + 5) = 3\) means \(2^3 = x + 5\)
Step 2: Evaluate and solve
\(8 = x + 5\)
\(x = 3\)
Step 3: Check domain
For \(\log_2(x + 5)\) to be defined, we need \(x + 5 > 0\)
Since \(3 + 5 = 8 > 0\), our solution is valid
✅ Answer: x = 3
📌 Paper 2 Questions (Calculator Allowed) – 3 Questions
Paper 2 – Q1. Use the change of base formula to evaluate \(\log_7 50\) correct to 3 significant figures.
[4 marks]
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Solution:
Step 1: Apply change of base formula
\(\log_7 50 = \frac{\log 50}{\log 7}\) (using base 10)
Step 2: Calculate using calculator
\(\log 50 = 1.6990\) and \(\log 7 = 0.8451\)
Step 3: Divide
\(\log_7 50 = \frac{1.6990}{0.8451} = 2.0103…\)
Step 4: Round to 3 significant figures
✅ Answer: 2.01
Paper 2 – Q2. Solve for x: \(3^{2x} = 5^{x+1}\) (Give your answer correct to 3 decimal places)
[5 marks]
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Solution:
Step 1: Take logarithm of both sides
\(\log(3^{2x}) = \log(5^{x+1})\)
Step 2: Apply power law
\(2x \log 3 = (x+1) \log 5\)
Step 3: Expand and collect x terms
\(2x \log 3 = x \log 5 + \log 5\)
\(2x \log 3 – x \log 5 = \log 5\)
\(x(2 \log 3 – \log 5) = \log 5\)
Step 4: Solve for x using calculator
\(x = \frac{\log 5}{2 \log 3 – \log 5} = \frac{0.6990}{2(0.4771) – 0.6990} = \frac{0.6990}{0.2552} = 2.740\)
✅ Answer: x = 2.740
Paper 2 – Q3. The pH of a solution is given by \(pH = -\log[H^+]\), where \([H^+]\) is the hydrogen ion concentration in mol/L.
(a) Find the pH of a solution with \([H^+] = 2.5 \times 10^{-4}\) mol/L. [2 marks]
(b) A solution has pH = 8.3. Find its hydrogen ion concentration. [3 marks]
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Solution:
(a) Finding pH:
Given: \([H^+] = 2.5 \times 10^{-4}\) mol/L
\(pH = -\log(2.5 \times 10^{-4})\)
Using calculator: \(pH = -(-3.602) = 3.602\)
Rounded: pH = 3.60
(b) Finding hydrogen ion concentration:
Given: pH = 8.3
\(8.3 = -\log[H^+]\)
\(\log[H^+] = -8.3\)
\([H^+] = 10^{-8.3}\)
Using calculator: \([H^+] = 5.01 \times 10^{-9}\) mol/L
✅ Answer: (a) pH = 3.60; (b) \([H^+] = 5.01 \times 10^{-9}\) mol/L
📌 Paper 3 Questions (Extended Response) – 3 Questions
Paper 3 – Q1. Given that \(\log_a 2 = p\) and \(\log_a 3 = q\), express the following in terms of p and q:
(a) \(\log_a 6\) [2 marks]
(b) \(\log_a \frac{4}{9}\) [3 marks]
(c) \(\log_a \sqrt{18}\) [4 marks]
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Complete Solution:
(a) Finding \(\log_a 6\):
Since \(6 = 2 \times 3\):
\(\log_a 6 = \log_a(2 \times 3) = \log_a 2 + \log_a 3 = p + q\)
(b) Finding \(\log_a \frac{4}{9}\):
Since \(4 = 2^2\) and \(9 = 3^2\):
\(\log_a \frac{4}{9} = \log_a \frac{2^2}{3^2} = \log_a 2^2 – \log_a 3^2\)
\(= 2\log_a 2 – 2\log_a 3 = 2p – 2q\)
(c) Finding \(\log_a \sqrt{18}\):
First simplify: \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} = 3 \times 2^{1/2}\)
\(\log_a \sqrt{18} = \log_a(3 \times 2^{1/2})\)
\(= \log_a 3 + \log_a 2^{1/2}\)
\(= \log_a 3 + \frac{1}{2}\log_a 2\)
\(= q + \frac{1}{2}p\)
✅ Final Answers:
(a) \(\log_a 6 = p + q\)
(b) \(\log_a \frac{4}{9} = 2p – 2q\)
(c) \(\log_a \sqrt{18} = q + \frac{1}{2}p\)
Paper 3 – Q2. The magnitude of an earthquake is given by the Richter scale formula: \(M = \log\left(\frac{A}{A_0}\right)\), where A is the amplitude of seismic waves and \(A_0\) is a reference amplitude.
(a) An earthquake has amplitude 50,000 times the reference amplitude. Find its magnitude. [2 marks]
(b) How many times greater is the amplitude of a magnitude 7 earthquake compared to a magnitude 4 earthquake? [3 marks]
(c) Two earthquakes occur with magnitudes 6.2 and 8.1 respectively. Express the ratio of their amplitudes in the form \(10^k\) where k is exact. [4 marks]
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Complete Solution:
(a) Finding magnitude:
Given: \(A = 50,000 A_0\)
\(M = \log\left(\frac{50,000 A_0}{A_0}\right) = \log(50,000)\)
\(M = \log(5 \times 10^4) = \log 5 + \log 10^4 = \log 5 + 4\)
Using calculator: \(M = 0.699 + 4 = 4.699\)
Magnitude ≈ 4.7
(b) Comparing magnitude 7 and 4:
For magnitude 7: \(7 = \log\left(\frac{A_7}{A_0}\right)\), so \(A_7 = 10^7 A_0\)
For magnitude 4: \(4 = \log\left(\frac{A_4}{A_0}\right)\), so \(A_4 = 10^4 A_0\)
Ratio: \(\frac{A_7}{A_4} = \frac{10^7 A_0}{10^4 A_0} = 10^3 = 1000\)
The magnitude 7 earthquake has 1000 times greater amplitude
(c) Ratio of amplitudes for magnitudes 6.2 and 8.1:
For magnitude 6.2: \(A_1 = 10^{6.2} A_0\)
For magnitude 8.1: \(A_2 = 10^{8.1} A_0\)
Ratio: \(\frac{A_2}{A_1} = \frac{10^{8.1} A_0}{10^{6.2} A_0} = 10^{8.1-6.2} = 10^{1.9}\)
✅ Final Answers:
(a) Magnitude = 4.7
(b) 1000 times greater
(c) \(10^{1.9}\)
Paper 3 – Q3. A radioactive substance decays according to the formula \(N(t) = N_0 e^{-\lambda t}\), where \(N(t)\) is the amount remaining after time t, \(N_0\) is the initial amount, and \(\lambda\) is the decay constant.
(a) A sample has a half-life of 20 years. Find the decay constant \(\lambda\). [3 marks]
(b) Starting with 100g, how much remains after 50 years? [2 marks]
(c) After how many years will only 10g remain? [4 marks]
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Complete Solution:
(a) Finding decay constant:
At half-life (t = 20), \(N(20) = \frac{N_0}{2}\)
\(\frac{N_0}{2} = N_0 e^{-20\lambda}\)
\(\frac{1}{2} = e^{-20\lambda}\)
Taking natural log: \(\ln\left(\frac{1}{2}\right) = -20\lambda\)
\(-\ln 2 = -20\lambda\)
\(\lambda = \frac{\ln 2}{20} = \frac{0.693}{20} = 0.0347\) per year
(b) Amount after 50 years:
Using \(N(t) = 100e^{-0.0347t}\) with t = 50:
\(N(50) = 100e^{-0.0347 \times 50} = 100e^{-1.735}\)
Using calculator: \(N(50) = 100 \times 0.176 = 17.6\) g
(c) Time for 10g to remain:
\(10 = 100e^{-0.0347t}\)
\(0.1 = e^{-0.0347t}\)
Taking natural log: \(\ln(0.1) = -0.0347t\)
\(-2.303 = -0.0347t\)
\(t = \frac{2.303}{0.0347} = 66.4\) years
✅ Final Answers:
(a) \(\lambda = 0.0347\) per year
(b) 17.6 g remains
(c) 66.4 years