IBDP Maths AA SL 1.7 Laws of exponents with rational exponents AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Rewrite \( 4^{x – 1} = (2^2)^{x – 1} = 2^{2x – 2} \).
Equation becomes: \( 2^{2x – 2} = 2^x + 8 \).
Multiply through by 4: \( 2^{2x} = 4 \cdot 2^x + 32 \).
Let \( y = 2^x \): \( y^2 – 4y – 32 = 0 \).
Solve: \( (y – 8)(y + 4) = 0 \implies y = 8 \) or \( y = -4 \).
Since \( 2^x > 0 \), discard \( y = -4 \).
For \( y = 8 \): \( 2^x = 8 = 2^3 \implies x = 3 \).
Markscheme
\( 4^{x – 1} = 2^{2x – 2} \quad \mathbf{(M1)} \).
\( 2^{2x – 2} = 2^x + 8 \quad \mathbf{(A1)} \).
\( 2^{2x} – 4 \cdot 2^x – 32 = 0 \quad \mathbf{A1} \).
\( (2^x – 8)(2^x + 4) = 0 \quad \mathbf{(M1)} \).
\( 2^x = 8 \implies x = 3 \quad \mathbf{A1} \).
Notes: Do not award final A1 if more than 1 solution is given.
[5 marks]
▶️ Answer/Explanation
Use change of base: \( \log_k m = \frac{\log m}{\log k} \).
\( a = \frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3} \times \frac{\log 5}{\log 4} \times \ldots \times \frac{\log 32}{\log 31} \).
Cancel terms: \( a = \frac{\log 32}{\log 2} \).
Since \( 32 = 2^5 \), \( \log 32 = 5 \log 2 \), so \( a = \frac{5 \log 2}{\log 2} = 5 \).
Thus, \( a = 5 \).
Markscheme
\( \frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3} \times \ldots \times \frac{\log 32}{\log 31} \quad \mathbf{M1A1} \).
\( = \frac{\log 32}{\log 2} \quad \mathbf{A1} \).
\( \log 32 = 5 \log 2 \), so \( \frac{5 \log 2}{\log 2} = 5 \quad \mathbf{(M1)} \).
\( a = 5 \quad \mathbf{A1} \).
Note: Accept if done in a specific base, e.g., \( \log_2 x \).
[5 marks]
▶️ Answer/Explanation
Take natural logarithms: \( \ln(8^{x – 1}) = \ln(6^{3x}) \).
Simplify: \( (x – 1) \ln 8 = 3x \ln 6 \).
Since \( 8 = 2^3 \), \( \ln 8 = 3 \ln 2 \); and \( 6 = 2 \cdot 3 \), \( \ln 6 = \ln 2 + \ln 3 \).
Equation becomes: \( 3(x – 1) \ln 2 = 3x (\ln 2 + \ln 3) \).
Divide by 3: \( (x – 1) \ln 2 = x (\ln 2 + \ln 3) \).
Expand and solve: \( x \ln 2 – \ln 2 = x \ln 2 + x \ln 3 \implies -\ln 2 = x \ln 3 \).
Thus, \( x = -\frac{\ln 2}{\ln 3} \).
Markscheme
METHOD 1
\( 8^{x – 1} = 2^{3(x – 1)} \), \( 6^{3x} = (2 \cdot 3)^{3x} \quad \mathbf{M1} \).
\( 2^{-3} = 3^{3x} \quad \mathbf{A1} \).
\( \ln(2^{-3}) = \ln(3^{3x}) \implies -3 \ln 2 = 3x \ln 3 \quad \mathbf{(M1)(A1)} \).
\( x = -\frac{\ln 2}{\ln 3} \quad \mathbf{A1} \).
METHOD 2
\( \ln(8^{x – 1}) = \ln(6^{3x}) \quad \mathbf{(M1)} \).
\( (x – 1) \ln 8 = 3x \ln 6 \implies 3(x – 1) \ln 2 = 3x (\ln 2 + \ln 3) \quad \mathbf{M1A1} \).
\( -\ln 2 = x \ln 3 \implies x = -\frac{\ln 2}{\ln 3} \quad \mathbf{(A1)(A1)} \).
METHOD 3
\( \ln(8^{x – 1}) = \ln(6^{3x}) \implies (x – 1) \ln 8 = 3x \ln 6 \quad \mathbf{(M1)(A1)} \).
\( x = \frac{\ln 8}{\ln 8 – 3 \ln 6} \quad \mathbf{A1} \).
\( \ln 8 = 3 \ln 2 \), \( \ln 6 = \ln 2 + \ln 3 \), so \( x = -\frac{\ln 2}{\ln 3} \quad \mathbf{(M1)(A1)} \).
[5 marks]
▶️ Answer/Explanation
Rewrite \( \log_9 6 \): \( \log_9 6 = \frac{\log_3 6}{\log_3 9} = \frac{\log_3 6}{2} \), so \( 10 \log_9 6 = 5 \log_3 6 \).
Since \( 6 = 2 \cdot 3 \), \( \log_3 6 = \log_3 2 + \log_3 3 = \log_3 2 + 1 \).
Thus, \( 5 \log_3 6 = 5 (\log_3 2 + 1) = 5 \log_3 2 + 5 \).
Equation becomes: \( m – n \log_3 2 = 5 + 5 \log_3 2 \).
Equate coefficients: \( m = 5 \), \( -n = 5 \implies n = -5 \).
Thus, \( m = 5 \), \( n = -5 \).
Markscheme
METHOD 1
\( 10 \log_9 6 = 5 \log_3 6 \quad \mathbf{M1} \).
\( m – n \log_3 2 = \log_3 (6^5 2^n) \quad \mathbf{(M1)} \).
\( 3^m 2^{-n} = 6^5 = 3^5 \cdot 2^5 \quad \mathbf{(M1)} \).
\( m = 5 \), \( n = -5 \quad \mathbf{A1} \).
Note: First M1 for correct change of base, second M1 for single logarithm, third M1 for writing 6 as \( 2 \cdot 3 \).
METHOD 2
\( 10 \log_9 6 = 5 \log_3 6 \quad \mathbf{M1} \).
\( 5 \log_3 6 = 5 (\log_3 3 + \log_3 2) \quad \mathbf{(M1)} \).
\( = 5 + 5 \log_3 2 \quad \mathbf{(M1)} \).
\( m = 5 \), \( n = -5 \quad \mathbf{A1} \).
Note: First M1 for correct change of base, second M1 for writing 6 as \( 2 \cdot 3 \), third M1 for eliminating \( \log_3 3 \).
[4 marks]
▶️ Answer/Explanation
Rewrite: \( 4^x = (2^2)^x = 2^{2x} \), \( 2^{x + 2} = 2^x \cdot 2^2 = 4 \cdot 2^x \).
Equation becomes: \( 2^{2x} + 4 \cdot 2^x – 3 = 0 \).
Let \( y = 2^x \): \( y^2 + 4y – 3 = 0 \).
Solve: \( y = \frac{-4 \pm \sqrt{16 + 12}}{2} = \frac{-4 \pm \sqrt{28}}{2} = -2 \pm \sqrt{7} \).
Since \( 2^x > 0 \), discard \( y = -2 – \sqrt{7} \).
Take \( y = -2 + \sqrt{7} \): \( 2^x = -2 + \sqrt{7} \implies x = \frac{\ln(-2 + \sqrt{7})}{\ln 2} \).
Markscheme
Attempt to form quadratic in \( 2^x \quad \mathbf{M1} \).
\( (2^x)^2 + 4 \cdot 2^x – 3 = 0 \quad \mathbf{A1} \).
\( 2^x = \frac{-4 \pm \sqrt{16 + 12}}{2} = -2 \pm \sqrt{7} \quad \mathbf{M1} \).
\( 2^x = -2 + \sqrt{7} \) (discard \( -2 – \sqrt{7} < 0 \)) \quad \mathbf{R1} \).
\( x = \frac{\ln(-2 + \sqrt{7})}{\ln 2} \quad \mathbf{A1} \).
Note: Award R0 A1 if final answer is \( x = \log_2(-2 + \sqrt{7}) \).
[5 marks]