IB Mathematics SL 1.5 Laws of exponents and logarithms AA SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Note: Answers to parts (b) and (c) must be given correct to two decimal places, or the final A1 is not awarded.
Part (a)
Given: Intensity of \( S_1 \), \( I_1 = 10^{-6} \) units.
\( S_2 \) has twice the intensity: \( I_2 = 2 \times 10^{-6} \) units (A1 N1)
[1 mark]
Part (b)
Loudness formula: \( L = 10 \log_{10} (I \times 10^{12}) \).
Substitute \( I_2 = 2 \times 10^{-6} \):
\( L_2 = 10 \log_{10} ((2 \times 10^{-6}) \times 10^{12}) = 10 \log_{10} (2 \times 10^6) \) (M1)
Simplify: \( \log_{10} (2 \times 10^6) = \log_{10} 2 + \log_{10} (10^6) \approx 0.3010 + 6 = 6.3010 \),
\( L_2 = 10 \times 6.3010 \approx 63.010 \).
Correct to 2 decimal places: 63.01 decibels (A1 N2)
[2 marks]
Part (c)
Method 1: Direct Logarithmic Solution
Given: \( L = 115 \), formula: \( L = 10 \log_{10} (I \times 10^{12}) \).
Substitute: \( 115 = 10 \log_{10} (I \times 10^{12}) \) (M1)
Solve: \( \log_{10} (I \times 10^{12}) = \frac{115}{10} = 11.5 \),
\( I \times 10^{12} = 10^{11.5} \),
\( I = \frac{10^{11.5}}{10^{12}} = 10^{11.5 – 12} = 10^{-0.5} \approx 0.316227766 \).
Correct to 2 decimal places: 0.32 units (A1 A1 N2)
Method 2: Using S1’s Loudness
Given: \( L_1 = 60 \), \( I_1 = 10^{-6} \), so \( 60 = 10 \log_{10} (10^{-6} \times 10^{12}) = 10 \log_{10} (10^6) \).
For thunder: \( L = 115 \). Loudness difference: \( 115 – 60 = 55 \).
Since \( L_2 – L_1 = 10 \log_{10} \left(\frac{I_2}{I_1}\right) \),
\( 55 = 10 \log_{10} \left(\frac{I_2}{10^{-6}}\right) \),
\( \log_{10} \left(\frac{I_2}{10^{-6}}\right) = 5.5 \),
\( \frac{I_2}{10^{-6}} = 10^{5.5} \), \( I_2 = 10^{5.5} \times 10^{-6} = 10^{-0.5} \approx 0.316227766 \) (M1)
Correct to 2 decimal places: 0.32 units (A1 A1 N2)
[3 marks]
Total [6 marks]