IB Mathematics SL 1.5 Laws of exponents AI SL Paper 1- Exam Style Questions- New Syllabus
Question
The loudness \(L\) (in dB) is given by \(L = 10 \log_{10}\!\left(\frac{I}{I_0}\right),\qquad I_0 = 10^{-12}\,\text{W m}^{-2}.\)
(a) The intensity on a busy street is \(I = 10^{-5}\,\text{W m}^{-2}\). Calculate \(L\).
(b) A jet‑engine sound reaches \(L = 185\) dB. Determine \(I\) in the form \(a \times 10^k\) with \(1 \le a < 10\), \(k \in \mathbb{Z}\).
(b) A jet‑engine sound reaches \(L = 185\) dB. Determine \(I\) in the form \(a \times 10^k\) with \(1 \le a < 10\), \(k \in \mathbb{Z}\).
Most‑appropriate topic codes (IB Mathematics: Applications and Interpretation 2025):
• SL 1.5: Introduction to logarithms with base 10 and e — parts (a), (b)
• SL 2.5: Modelling with exponential and logarithmic functions — contextual application
• SL 2.5: Modelling with exponential and logarithmic functions — contextual application
▶️ Answer/Explanation
(a)
\(L = 10 \log_{10}\!\left(\frac{10^{-5}}{10^{-12}}\right) = 10 \log_{10}\!\bigl(10^{7}\bigr) = 10 \times 7 = 70\)
\(\boxed{70\ \text{dB}}\)
(b)
\(185 = 10 \log_{10}\!\left(\frac{I}{10^{-12}}\right)\)
Divide by 10: \(18.5 = \log_{10}\!\left(\frac{I}{10^{-12}}\right)\)
Convert from logarithmic form: \(\frac{I}{10^{-12}} = 10^{18.5}\)
\(I = 10^{-12} \times 10^{18.5} = 10^{6.5}\)
Now express \(10^{6.5}\) in the required form \(a \times 10^k\):
\(10^{6.5} = 10^{0.5} \times 10^{6} = \sqrt{10} \times 10^{6} \approx 3.16227766 \times 10^{6}\)
\(\boxed{3.16 \times 10^{6}\ \text{W m}^{-2}}\)