IB Mathematics SL 1.5 Laws of exponents AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Sophie studies earthquakes in a region where the strength is measured on the Richter magnitude scale, with values typically between 0 and 8, where 8 is the most severe.
The Gutenberg–Richter equation gives the average number of earthquakes per year, \( N \), which have a magnitude of at least \( M \). For this region, the equation is \( \log_{10} N = a – M \), for some \( a \in \mathbb{R} \).
This region has an average of 100 earthquakes per year with a magnitude of at least 3.
(a) Find the value of \( a \). [2]
The equation for this region can also be written as \( N = \frac{b}{10^M} \).
(b) Find the value of \( b \). [2]
(c) Given \( 0 < M < 8 \), find the range for \( N \). [2]
The expected length of time, in years, between earthquakes with a magnitude of at least \( M \) is \( \frac{1}{N} \).
Within this region, the most severe earthquake recorded had a magnitude of 7.2.
(d) Find the expected length of time between this earthquake and the next earthquake of at least this magnitude. Give your answer to the nearest year. [2]
▶️ Answer/Explanation
Markscheme
(a)
Given \( N = 100 \) when \( M = 3 \):
\[ \begin{aligned} \log_{10} 100 &= a – 3 \\ 2 &= a – 3 \\ a &= 5 \end{aligned} \] (M1) A1
[2 marks]
Given \( N = 100 \) when \( M = 3 \):
\[ \begin{aligned} \log_{10} 100 &= a – 3 \\ 2 &= a – 3 \\ a &= 5 \end{aligned} \] (M1) A1
[2 marks]
(b)
Rewrite \( \log_{10} N = 5 – M \):
\[ \begin{aligned} N &= 10^{5 – M} \\ &= \frac{10^5}{10^M} \end{aligned} \]
Thus, \( b = 10^5 = 100000 \). (M1) A1
Alternatively, use \( N = 100 \), \( M = 3 \):
\[ \begin{aligned} 100 &= \frac{b}{10^3} \\ b &= 100 \times 10^3 = 100000 \end{aligned} \] [2 marks]
Rewrite \( \log_{10} N = 5 – M \):
\[ \begin{aligned} N &= 10^{5 – M} \\ &= \frac{10^5}{10^M} \end{aligned} \]
Thus, \( b = 10^5 = 100000 \). (M1) A1
Alternatively, use \( N = 100 \), \( M = 3 \):
\[ \begin{aligned} 100 &= \frac{b}{10^3} \\ b &= 100 \times 10^3 = 100000 \end{aligned} \] [2 marks]
(c)
For \( 0 < M < 8 \), \( N = 10^{5 – M} \):
\[ \begin{aligned} \text{When } M = 0, \quad N &= 10^5 = 100000 \\ \text{When } M \to 8, \quad N &\to 10^{5 – 8} = 10^{-3} = 0.001 \end{aligned} \]
Range: \( 0.001 < N < 100000 \). A1 A1
[2 marks]
For \( 0 < M < 8 \), \( N = 10^{5 – M} \):
\[ \begin{aligned} \text{When } M = 0, \quad N &= 10^5 = 100000 \\ \text{When } M \to 8, \quad N &\to 10^{5 – 8} = 10^{-3} = 0.001 \end{aligned} \]
Range: \( 0.001 < N < 100000 \). A1 A1
[2 marks]
(d)
For \( M = 7.2 \), calculate \( N \):
\[ \begin{aligned} N &= 10^{5 – 7.2} = 10^{-2.2} \approx 0.0063095 \end{aligned} \]
Expected time:
\[ \begin{aligned} \frac{1}{N} &\approx \frac{1}{0.0063095} \approx 158.4893 \end{aligned} \]
To the nearest year: 158 years. (M1) A1
[2 marks]
For \( M = 7.2 \), calculate \( N \):
\[ \begin{aligned} N &= 10^{5 – 7.2} = 10^{-2.2} \approx 0.0063095 \end{aligned} \]
Expected time:
\[ \begin{aligned} \frac{1}{N} &\approx \frac{1}{0.0063095} \approx 158.4893 \end{aligned} \]
To the nearest year: 158 years. (M1) A1
[2 marks]
Total Marks: 8