IB Mathematics SL 1.5 Laws of exponents and logarithms AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a) [2 marks]
Method 1: Given \( \log_{10}a = \frac{1}{3} \), use the logarithm property: \( \log_b\left(\frac{1}{x}\right) = -\log_b x \).
\[ \log_{10}\left(\frac{1}{a}\right) = -\log_{10}a = -\frac{1}{3} \]
Method 2: Since \( \log_{10}a = \frac{1}{3} \), we have \( a = 10^{\frac{1}{3}} \).
\[ \frac{1}{a} = \frac{1}{10^{\frac{1}{3}}} = 10^{-\frac{1}{3}} \]
\[ \log_{10}\left(10^{-\frac{1}{3}}\right) = -\frac{1}{3} \]
Answer: \( -\frac{1}{3} \)
Part (b) [3 marks]
Method 1: Given \( \log_{10}a = \frac{1}{3} \), and \( 1000 = 10^3 \), use the change of base formula: \( \log_b x = \frac{\log_k x}{\log_k b} \).
\[ \log_{1000}a = \frac{\log_{10}a}{\log_{10}1000} = \frac{\frac{1}{3}}{3} = \frac{1}{9} \]
Method 2: Let \( \log_{1000}a = x \). Then \( 1000^x = a \).
Since \( 1000 = 10^3 \) and \( a = 10^{\frac{1}{3}} \):
\[ (10^3)^x = 10^{\frac{1}{3}} \]
\[ 10^{3x} = 10^{\frac{1}{3}} \]
Equate exponents: \( 3x = \frac{1}{3} \), so \( x = \frac{1}{9} \).
Answer: \( \frac{1}{9} \)
▶️ Answer/Explanation
Part (a) [2 marks]
Method 1: Use the change of base formula: \( \log_b x = \frac{\log_k x}{\log_k b} \).
\[ \log_{r^2} x = \frac{\log_r x}{\log_r r^2} \]
Since \( \log_r r^2 = 2 \log_r r = 2 \):
\[ \log_{r^2} x = \frac{\log_r x}{2} \]
Method 2: Use the alternative form: \( \log_{r^2} x = \frac{1}{\log_x r^2} \).
\[ \log_x r^2 = 2 \log_x r \]
\[ \log_{r^2} x = \frac{1}{2 \log_x r} = \frac{\log_r x}{2} \]
Answer: \( \log_{r^2} x = \frac{1}{2} \log_r x \), as required.
Part (b) [5 marks]
Method 1: Given: \( \log_2 y + \log_4 x + \log_4 2x = 0 \).
Combine logarithmic terms: \( \log_4 x + \log_4 2x = \log_4 (x \cdot 2x) = \log_4 (2x^2) \).
Equation: \( \log_2 y + \log_4 2x^2 = 0 \).
Change base for \( \log_4 2x^2 \): Since \( 4 = 2^2 \), \( \log_4 2x^2 = \frac{\log_2 2x^2}{\log_2 4} = \frac{\log_2 2x^2}{2} \).
\[ \log_2 2x^2 = \log_2 2 + \log_2 x^2 = 1 + 2 \log_2 x \]
\[ \log_4 2x^2 = \frac{1 + 2 \log_2 x}{2} \]
Equation becomes: \( \log_2 y + \frac{1 + 2 \log_2 x}{2} = 0 \).
\[ \log_2 y = -\frac{1 + 2 \log_2 x}{2} = -\frac{1}{2} – \log_2 x \]
\[ \log_2 y = \log_2 \left( 2^{-1/2} x^{-1} \right) \]
\[ y = 2^{-1/2} x^{-1} = \frac{1}{\sqrt{2}} x^{-1} \]
Method 2: Convert all terms to base 2:
\[ \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2} \]
\[ \log_4 2x = \frac{\log_2 2x}{2} = \frac{\log_2 2 + \log_2 x}{2} = \frac{1 + \log_2 x}{2} \]
Equation: \( \log_2 y + \frac{\log_2 x}{2} + \frac{1 + \log_2 x}{2} = 0 \).
\[ \log_2 y + \frac{\log_2 x + 1 + \log_2 x}{2} = \log_2 y + \frac{1 + 2 \log_2 x}{2} = 0 \]
\[ \log_2 y = -\frac{1 + 2 \log_2 x}{2} = \log_2 \left( 2^{-1/2} x^{-1} \right) \]
\[ y = \frac{1}{\sqrt{2}} x^{-1} \]
Answer: \( y = \frac{1}{\sqrt{2}} x^{-1} \), where \( p = \frac{1}{\sqrt{2}} \), \( q = -1 \).
Part (c) [5 marks]
From part (b), \( y = \frac{1}{\sqrt{2}} x^{-1} \).
The area of region \( R \) is given by: \( \int_1^\alpha \frac{1}{\sqrt{2}} x^{-1} \, dx = \sqrt{2} \).
Compute the integral: \( \int \frac{1}{\sqrt{2}} x^{-1} \, dx = \frac{1}{\sqrt{2}} \int x^{-1} \, dx = \frac{1}{\sqrt{2}} \ln x \).
Evaluate: \( \left[ \frac{1}{\sqrt{2}} \ln x \right]_1^\alpha = \frac{1}{\sqrt{2}} (\ln \alpha – \ln 1) = \frac{1}{\sqrt{2}} \ln \alpha \).
Set equal to the area: \( \frac{1}{\sqrt{2}} \ln \alpha = \sqrt{2} \).
Solve: \( \ln \alpha = \sqrt{2} \cdot \sqrt{2} = 2 \).
\[ \alpha = e^2 \]
Answer: \( \alpha = e^2 \).