LAWS OF EXPONENTS & RATIONAL LOGARITHMS
This question bank contains 17 questions covering advanced laws of exponents and rational logarithmic operations, distributed across different paper types according to IB AAHL curriculum standards.
📌 Multiple Choice Questions (3 Questions)
MCQ 1. What is the value of \(32^{3/5}\)?
A) 6 B) 8 C) 12 D) 16
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Solution:
Method 1: \(32^{3/5} = (\sqrt[5]{32})^3 = 2^3 = 8\)
Method 2: \(32^{3/5} = \sqrt[5]{32^3} = \sqrt[5]{32768} = 8\)
Verification: Since \(32 = 2^5\), we have \(32^{3/5} = (2^5)^{3/5} = 2^3 = 8\)
✅ Answer: B) 8
MCQ 2. If \(2^{3x} = 8^{x-1}\), what is the value of x?
A) -3 B) -1 C) 1 D) 3
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Solution:
Express both sides with base 2: \(8 = 2^3\)
\(2^{3x} = (2^3)^{x-1} = 2^{3(x-1)} = 2^{3x-3}\)
Since bases are equal: \(3x = 3x – 3\)
This gives us: \(0 = -3\), which is impossible!
Let me recalculate: \(3x = 3x – 3\) means no solution unless I made an error.
Actually: \(2^{3x} = 2^{3x-3}\) implies \(3x = 3x – 3\) only if the bases are different.
Wait, let me solve this correctly: \(3x = 3(x-1) = 3x – 3\) gives \(3x = 3x – 3\), so \(0 = -3\).
This suggests there’s no solution, but let me check the options by substitution…
For x = -1: LHS = \(2^{3(-1)} = 2^{-3} = 1/8\); RHS = \(8^{-1-1} = 8^{-2} = 1/64\) ≠ 1/8
Actually, I think there’s an error in my setup. Let me be more careful:
\(2^{3x} = 8^{x-1}\) becomes \(2^{3x} = (2^3)^{x-1} = 2^{3(x-1)}\)
So: \(3x = 3(x-1) = 3x – 3\), which gives \(0 = -3\) (impossible)
⚠️ This equation has no solution. There may be an error in the question.
MCQ 3. What is the domain of \(f(x) = \log_3(2x – 6)\)?
A) \(x > 0\) B) \(x > 3\) C) \(x > 6\) D) \(x \geq 3\)
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Solution:
For the logarithm to be defined, the argument must be positive:
\(2x – 6 > 0\)
\(2x > 6\)
\(x > 3\)
✅ Answer: B) \(x > 3\)
📌 Paper 1 Questions (No Calculator) – 6 Questions
Paper 1 – Q1. Simplify: \(25^{3/2} – 8^{2/3} + 27^{-1/3}\)
[4 marks]
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Solution:
Step 1: Evaluate \(25^{3/2}\)
\(25^{3/2} = (\sqrt{25})^3 = 5^3 = 125\)
Step 2: Evaluate \(8^{2/3}\)
\(8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4\)
Step 3: Evaluate \(27^{-1/3}\)
\(27^{-1/3} = \frac{1}{27^{1/3}} = \frac{1}{\sqrt[3]{27}} = \frac{1}{3}\)
Step 4: Combine
\(125 – 4 + \frac{1}{3} = 121 + \frac{1}{3} = \frac{363 + 1}{3} = \frac{364}{3}\)
✅ Answer: \(\frac{364}{3}\) or \(121\frac{1}{3}\)
Paper 1 – Q2. Solve for x: \(3^{2x+1} = 9^{x-2}\)
[4 marks]
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Solution:
Step 1: Express both sides with base 3
Since \(9 = 3^2\): \(3^{2x+1} = (3^2)^{x-2}\)
Step 2: Simplify the right side
\(3^{2x+1} = 3^{2(x-2)} = 3^{2x-4}\)
Step 3: Equate exponents
\(2x + 1 = 2x – 4\)
Step 4: Solve
\(1 = -4\)
❌ No solution (the equation is inconsistent)
Note: This suggests the original equation has no solution.
Paper 1 – Q3. Solve for x: \(\log_2(x + 4) = 3\)
[3 marks]
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Solution:
Step 1: Convert to exponential form
\(\log_2(x + 4) = 3\) means \(2^3 = x + 4\)
Step 2: Solve for x
\(8 = x + 4\)
\(x = 4\)
Step 3: Check domain
Need \(x + 4 > 0\): \(4 + 4 = 8 > 0\) ✓
✅ Answer: x = 4
Paper 1 – Q4. Express \(x^{-3/4}\) in radical form and find its value when x = 16.
[4 marks]
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Solution:
Step 1: Express in radical form
\(x^{-3/4} = \frac{1}{x^{3/4}} = \frac{1}{(\sqrt[4]{x})^3}\) or \(\frac{1}{\sqrt[4]{x^3}}\)
Step 2: Substitute x = 16
\(16^{-3/4} = \frac{1}{16^{3/4}} = \frac{1}{(\sqrt[4]{16})^3}\)
Step 3: Evaluate
\(\sqrt[4]{16} = 2\) (since \(2^4 = 16\))
Therefore: \(\frac{1}{2^3} = \frac{1}{8}\)
✅ Answer: \(\frac{1}{(\sqrt[4]{x})^3}\); when x = 16, value is \(\frac{1}{8}\)
Paper 1 – Q5. Solve for x: \(\log_3(x) + \log_3(x – 2) = 1\)
[5 marks]
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Solution:
Step 1: Apply logarithm product law
\(\log_3[x(x-2)] = 1\)
Step 2: Convert to exponential form
\(x(x-2) = 3^1 = 3\)
Step 3: Expand and solve
\(x^2 – 2x = 3\)
\(x^2 – 2x – 3 = 0\)
\((x-3)(x+1) = 0\)
Step 4: Find solutions
\(x = 3\) or \(x = -1\)
Step 5: Check domain restrictions
Need \(x > 0\) and \(x – 2 > 0\) (so \(x > 2\))
\(x = 3\): 3 > 2 ✓
\(x = -1\): -1 < 2 ✗
✅ Answer: x = 3
Paper 1 – Q6. If \(a^{1/2} = 3\), find the value of \(a^{3/2}\).
[3 marks]
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Solution:
Method 1: Using exponent laws
\(a^{3/2} = a^{1/2 + 1} = a^{1/2} \times a^1\)
We need to find \(a\) first: \(a^{1/2} = 3\) means \(a = 3^2 = 9\)
Therefore: \(a^{3/2} = 9^{3/2} = (\sqrt{9})^3 = 3^3 = 27\)
Method 2: Direct approach
\(a^{3/2} = (a^{1/2})^3 = 3^3 = 27\)
✅ Answer: 27
📌 Paper 2 Questions (Calculator Allowed) – 3 Questions
Paper 2 – Q1. Solve for x: \(2^{3x} – 7 \cdot 2^x + 6 = 0\). Give your answers as exact values.
[6 marks]
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Solution:
Step 1: Make a substitution
Let \(u = 2^x\), then \(2^{3x} = (2^x)^3 = u^3\)
The equation becomes: \(u^3 – 7u + 6 = 0\)
Step 2: Factor the cubic
Try \(u = 1\): \(1 – 7 + 6 = 0\) ✓
So \((u – 1)\) is a factor: \(u^3 – 7u + 6 = (u – 1)(u^2 + u – 6)\)
Factor the quadratic: \(u^2 + u – 6 = (u + 3)(u – 2)\)
Step 3: Find values of u
\((u – 1)(u + 3)(u – 2) = 0\)
So \(u = 1\), \(u = -3\), or \(u = 2\)
Step 4: Substitute back
Case 1: \(2^x = 1 = 2^0\), so \(x = 0\)
Case 2: \(2^x = -3\) (impossible since \(2^x > 0\))
Case 3: \(2^x = 2 = 2^1\), so \(x = 1\)
✅ Answer: x = 0 and x = 1
Paper 2 – Q2. Use the change of base formula to solve: \(\log_5 x = \log_3 12\). Give your answer correct to 3 significant figures.
[5 marks]
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Solution:
Step 1: Apply change of base formula
\(\log_5 x = \log_3 12\) can be written as:
\(\frac{\log x}{\log 5} = \frac{\log 12}{\log 3}\)
Step 2: Solve for log x
\(\log x = \frac{\log 12 \times \log 5}{\log 3}\)
Step 3: Calculate using calculator
\(\log 12 = 1.0792\)
\(\log 5 = 0.6990\)
\(\log 3 = 0.4771\)
\(\log x = \frac{1.0792 \times 0.6990}{0.4771} = \frac{0.7544}{0.4771} = 1.581\)
Step 4: Find x
\(x = 10^{1.581} = 38.1\)
✅ Answer: x = 38.1
Paper 2 – Q3. A radioactive substance decays according to \(N(t) = N_0 e^{-0.0231t}\), where t is in years.
(a) What is the half-life of this substance? [3 marks]
(b) How long will it take for 90% of the substance to decay? [3 marks]
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Solution:
(a) Half-life calculation:
At half-life, \(N(t) = \frac{N_0}{2}\):
\(\frac{N_0}{2} = N_0 e^{-0.0231t}\)
\(\frac{1}{2} = e^{-0.0231t}\)
Take natural log: \(\ln(0.5) = -0.0231t\)
\(t = \frac{\ln(0.5)}{-0.0231} = \frac{-0.693}{-0.0231} = 30.0\) years
(b) Time for 90% to decay:
If 90% decays, 10% remains: \(N(t) = 0.1N_0\)
\(0.1N_0 = N_0 e^{-0.0231t}\)
\(0.1 = e^{-0.0231t}\)
Take natural log: \(\ln(0.1) = -0.0231t\)
\(t = \frac{\ln(0.1)}{-0.0231} = \frac{-2.303}{-0.0231} = 99.7\) years
✅ Answer: (a) 30.0 years; (b) 99.7 years
📌 Paper 3 Questions (Extended Response) – 5 Questions
Paper 3 – Q1. The population of a city grows according to \(P(t) = 100000(1.03)^t\), where t is years since 2000.
(a) What was the population in 2000? [1 mark]
(b) What is the annual growth rate? [1 mark]
(c) In which year will the population reach 150,000? [3 marks]
(d) How long does it take for the population to double? [3 marks]
(e) Express the model in the form \(P(t) = P_0 e^{kt}\) and find k. [3 marks]
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Complete solution:
(a) Population in 2000:
At t = 0: \(P(0) = 100000(1.03)^0 = 100000\)
(b) Annual growth rate:
From \(P(t) = P_0(1 + r)^t\), we have \(1 + r = 1.03\)
Therefore, \(r = 0.03 = 3\%\) per year
(c) When population reaches 150,000:
\(150000 = 100000(1.03)^t\)
\(1.5 = (1.03)^t\)
Take logarithm: \(\log(1.5) = t \log(1.03)\)
\(t = \frac{\log(1.5)}{\log(1.03)} = \frac{0.1761}{0.0128} = 13.7\) years
So in year 2000 + 13.7 = 2014 (approximately)
(d) Time to double:
\(200000 = 100000(1.03)^t\)
\(2 = (1.03)^t\)
\(t = \frac{\log(2)}{\log(1.03)} = \frac{0.3010}{0.0128} = 23.4\) years
(e) Converting to exponential form:
We want \(100000(1.03)^t = P_0 e^{kt}\)
Since \(P_0 = 100000\): \((1.03)^t = e^{kt}\)
Taking natural log: \(t \ln(1.03) = kt\)
Therefore: \(k = \ln(1.03) = 0.0296\)
Model: \(P(t) = 100000 e^{0.0296t}\)
✅ Final Answers:
(a) 100,000 people
(b) 3% per year
(c) 2014
(d) 23.4 years
(e) \(P(t) = 100000 e^{0.0296t}\), k = 0.0296
Paper 3 – Q2. Solve the system of equations:
\(2^x \cdot 3^y = 72\)
\(4^x \cdot 9^y = 5184\)
[8 marks]
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Solution:
Step 1: Rewrite with common bases
First equation: \(2^x \cdot 3^y = 72\)
Second equation: \((2^2)^x \cdot (3^2)^y = 5184\)
Simplify: \(2^{2x} \cdot 3^{2y} = 5184\)
Step 2: Square the first equation
\((2^x \cdot 3^y)^2 = 72^2\)
\(2^{2x} \cdot 3^{2y} = 5184\)
Step 3: Verify consistency
This matches our second equation, so the system is consistent.
Step 4: Solve the first equation
Factor 72: \(72 = 8 \times 9 = 2^3 \times 3^2\)
So \(2^x \cdot 3^y = 2^3 \cdot 3^2\)
Step 5: Compare exponents
Since the bases are coprime: \(x = 3\) and \(y = 2\)
Step 6: Verify in second equation
\(4^3 \cdot 9^2 = 64 \times 81 = 5184\) ✓
✅ Answer: x = 3, y = 2
Paper 3 – Q3. A drug concentration in the bloodstream follows \(C(t) = 20e^{-0.3t}\) mg/L, where t is hours after injection.
(a) What is the initial concentration? [1 mark]
(b) When will the concentration drop to 5 mg/L? [3 marks]
(c) What is the half-life of the drug? [3 marks]
(d) If the therapeutic window is between 2 mg/L and 15 mg/L, for how long is the drug in the therapeutic range? [4 marks]
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Complete solution:
(a) Initial concentration:
At t = 0: \(C(0) = 20e^{-0.3(0)} = 20e^0 = 20\) mg/L
(b) When concentration drops to 5 mg/L:
\(5 = 20e^{-0.3t}\)
\(0.25 = e^{-0.3t}\)
Take natural log: \(\ln(0.25) = -0.3t\)
\(t = \frac{\ln(0.25)}{-0.3} = \frac{-1.386}{-0.3} = 4.62\) hours
(c) Half-life:
When \(C(t) = 10\) mg/L (half of 20):
\(10 = 20e^{-0.3t}\)
\(0.5 = e^{-0.3t}\)
\(t = \frac{\ln(0.5)}{-0.3} = \frac{-0.693}{-0.3} = 2.31\) hours
(d) Therapeutic window (2 mg/L to 15 mg/L):
Upper limit (15 mg/L): \(15 = 20e^{-0.3t_1}\)
\(0.75 = e^{-0.3t_1}\)
\(t_1 = \frac{\ln(0.75)}{-0.3} = 0.959\) hours
Lower limit (2 mg/L): \(2 = 20e^{-0.3t_2}\)
\(0.1 = e^{-0.3t_2}\)
\(t_2 = \frac{\ln(0.1)}{-0.3} = 7.68\) hours
Duration in therapeutic range: \(t_2 – t_1 = 7.68 – 0.959 = 6.72\) hours
✅ Final Answers:
(a) 20 mg/L
(b) 4.62 hours
(c) 2.31 hours
(d) 6.72 hours
Paper 3 – Q4. Consider the function \(f(x) = 2^x + 2^{-x}\).
(a) Show that \(f(-x) = f(x)\). [2 marks]
(b) Find \(f(1)\) and \(f(2)\). [2 marks]
(c) Solve \(f(x) = 5\). [4 marks]
(d) Find the minimum value of \(f(x)\) and justify your answer. [4 marks]
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Complete solution:
(a) Showing f(-x) = f(x):
\(f(-x) = 2^{-x} + 2^{-(-x)} = 2^{-x} + 2^x = f(x)\)
Therefore, f(x) is an even function.
(b) Evaluating f(1) and f(2):
\(f(1) = 2^1 + 2^{-1} = 2 + 0.5 = 2.5\)
\(f(2) = 2^2 + 2^{-2} = 4 + 0.25 = 4.25\)
(c) Solving f(x) = 5:
\(2^x + 2^{-x} = 5\)
Let \(u = 2^x\), then \(2^{-x} = \frac{1}{u}\):
\(u + \frac{1}{u} = 5\)
Multiply by u: \(u^2 + 1 = 5u\)
\(u^2 – 5u + 1 = 0\)
Using quadratic formula:
\(u = \frac{5 \pm \sqrt{25-4}}{2} = \frac{5 \pm \sqrt{21}}{2}\)
Since \(u = 2^x > 0\), both solutions are valid:
\(2^x = \frac{5 + \sqrt{21}}{2}\) or \(2^x = \frac{5 – \sqrt{21}}{2}\)
Taking logarithms:
\(x = \log_2\left(\frac{5 + \sqrt{21}}{2}\right)\) or \(x = \log_2\left(\frac{5 – \sqrt{21}}{2}\right)\)
(d) Minimum value:
By AM-GM inequality: \(\frac{2^x + 2^{-x}}{2} \geq \sqrt{2^x \cdot 2^{-x}} = \sqrt{2^0} = 1\)
Therefore: \(2^x + 2^{-x} \geq 2\)
Equality occurs when \(2^x = 2^{-x}\), i.e., when \(x = 0\)
Check: \(f(0) = 2^0 + 2^0 = 1 + 1 = 2\)
✅ Final Answers:
(a) Proven: f(x) is even
(b) f(1) = 2.5, f(2) = 4.25
(c) \(x = \log_2\left(\frac{5 \pm \sqrt{21}}{2}\right)\)
(d) Minimum value is 2 at x = 0
Paper 3 – Q5. The magnitude of sound intensity is measured using \(L = 10\log\left(\frac{I}{I_0}\right)\) decibels, where \(I_0 = 10^{-12}\) W/m² is the reference intensity.
(a) A whisper has intensity \(I = 10^{-10}\) W/m². Find its sound level in decibels. [2 marks]
(b) A rock concert measures 110 dB. Find the intensity of sound. [3 marks]
(c) How many times more intense is the rock concert compared to the whisper? [2 marks]
(d) If two identical sound sources are combined, show that the total sound level increases by approximately 3 dB. [4 marks]
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Complete solution:
(a) Whisper sound level:
\(L = 10\log\left(\frac{10^{-10}}{10^{-12}}\right) = 10\log(10^2) = 10 \times 2 = 20\) dB
(b) Rock concert intensity:
\(110 = 10\log\left(\frac{I}{10^{-12}}\right)\)
\(11 = \log\left(\frac{I}{10^{-12}}\right)\)
\(10^{11} = \frac{I}{10^{-12}}\)
\(I = 10^{11} \times 10^{-12} = 10^{-1} = 0.1\) W/m²
(c) Intensity ratio:
Ratio = \(\frac{0.1}{10^{-10}} = \frac{10^{-1}}{10^{-10}} = 10^9\)
The rock concert is 1 billion times more intense.
(d) Combined sources:
Let each source have intensity I. Combined intensity = 2I
Original level: \(L_1 = 10\log\left(\frac{I}{I_0}\right)\)
Combined level: \(L_2 = 10\log\left(\frac{2I}{I_0}\right)\)
Difference:
\(L_2 – L_1 = 10\log\left(\frac{2I}{I_0}\right) – 10\log\left(\frac{I}{I_0}\right)\)
\(= 10\log\left(\frac{2I/I_0}{I/I_0}\right) = 10\log(2)\)
\(= 10 \times 0.301 = 3.01 ≈ 3\) dB
✅ Final Answers:
(a) 20 dB
(b) 0.1 W/m²
(c) 1 billion times
(d) Increase = 10log(2) ≈ 3 dB