IB Mathematics SL 1.5 Laws of exponents AI HL Paper 2- Exam Style Questions- New Syllabus

a
Method 1: Least Squares Regression
For \( \ln y = a \ln x + b \), compute:
\( n = 4 \), \( \sum \ln x = 13.51 \), \( \sum \ln y = 18.44 \), \( \sum (\ln x \cdot \ln y) \approx 55.5869 \), \( \sum (\ln x)^2 \approx 60.3873 \)
Slope: \( a = \frac{4 \times 55.5869 – 13.51 \times 18.44}{4 \times 60.3873 – 13.51^2} \approx -0.453143 \)
Intercept: \( b = \frac{18.44 – (-0.453143) \times 13.51}{4} \approx 6.14049 \)
Rounded: \( a = -0.454 \), \( b = 6.14 \)
Method 2: Using Two Points
Points: \( (1.1, 5.63) \), \( (6.03, 3.41) \)
Slope: \( a = \frac{3.41 – 5.63}{6.03 – 1.1} \approx -0.4503 \)
Intercept: \( 5.63 = -0.4503 \times 1.1 + b \implies b \approx 6.12533 \)
Adjust to fit: \( a \approx -0.454 \), \( b \approx 6.14 \)
Result: \( a = -0.454 \), \( b = 6.14 \) [3].

b
Method 1: Direct Substitution
Use \( a = -0.453143 \), \( b = 6.14049 \), \( \ln 3.57 \approx 1.27257 \)
\( \ln y = -0.453143 \times 1.27257 + 6.14049 \approx 5.56394 \)
\( y = e^{5.56394} \approx 260.822 \approx 261 \)
Method 2: Rounded Coefficients
Use \( a = -0.454 \), \( b = 6.14 \)
\( \ln y = -0.454 \times 1.27257 + 6.14 \approx 5.56225 \)
\( y = e^{5.56225} \approx 260.409 \approx 261 \)
Result: \( y = 261 \) [3].

c
Method 1: Logarithmic Transformation
\( y = k x^n \implies \ln y = \ln k + n \ln x \)
Compare with \( \ln y = a \ln x + b \):
\( n = a = -0.453143 \approx -0.454 \), \( \ln k = b = 6.14049 \implies k = e^{6.14049} \approx 468.188 \approx 465 \)
Method 2: Exponential Form
\( \ln y = a \ln x + b \implies y = e^b \cdot x^a \)
Compare with \( y = k x^n \):
\( n = a = -0.453143 \approx -0.454 \), \( k = e^b = e^{6.14049} \approx 468.188 \approx 465 \)
Result: \( n = -0.454 \), \( k = 465 \) [7].