IB Mathematics AHL 1.9 Laws of logarithms AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
a
Method 1: Change of Base
\( \log_{r^2} x = \frac{\log_r x}{\log_r (r^2)} \)
\( \log_r (r^2) = 2 \log_r r = 2 \)
\( \log_{r^2} x = \frac{\log_r x}{2} = \frac{1}{2} \log_r x \)
Method 2: Inverse Base
\( \log_{r^2} x = \frac{1}{\log_x (r^2)} = \frac{1}{2 \log_x r} = \frac{1}{2} \log_r x \)
Result: \( \log_{r^2} x = \frac{1}{2} \log_r x \) [2].
b
Method 1: Combine Logarithms
\( \log_4 x + \log_4 (2x) = \log_4 (2x^2) \)
\( \log_2 y + \frac{\log_2 (2x^2)}{2} = 0 \)
\( \log_2 y = -\frac{1 + 2 \log_2 x}{2} = \log_2 \left( \frac{1}{\sqrt{2} x} \right) \)
\( y = \frac{1}{\sqrt{2}} x^{-1} \)
Method 2: Unified Base
\( \log_4 x = \frac{\log_2 x}{2} \), \( \log_4 (2x) = \frac{1 + \log_2 x}{2} \)
\( \log_2 y + \frac{\log_2 x + 1 + \log_2 x}{2} = 0 \)
\( \log_2 y = \log_2 \left( \frac{1}{\sqrt{2} x} \right) \)
\( y = \frac{1}{\sqrt{2}} x^{-1} \)
Result: \( y = \frac{1}{\sqrt{2}} x^{-1} \) [5].
c
Area of \( R \): \( \int_1^\alpha \frac{1}{\sqrt{2}} x^{-1} \, dx \)
\( = \left[ \frac{1}{\sqrt{2}} \ln x \right]_1^\alpha = \frac{1}{\sqrt{2}} \ln \alpha \)
\( \frac{1}{\sqrt{2}} \ln \alpha = \sqrt{2} \)
\( \ln \alpha = 2 \implies \alpha = e^2 \)
Result: \( \alpha = e^2 \) [5].