IBDP MAA SL 1.5: Laws of exponents with integer HL Paper 1- Exam Questions
▶️ Answer/Explanation
Rewrite \( 9^x = (3^x)^2 \):
\( 3 (3^x)^2 + 5 (3^x) – 2 = 0 \).
Let \( t = 3^x \):
\( 3t^2 + 5t – 2 = 0 \).
Factorize: \( (3t – 1)(t + 2) = 0 \).
Solve: \( t = \frac{1}{3} \) or \( t = -2 \).
Since \( 3^x > 0 \), discard \( t = -2 \).
For \( t = \frac{1}{3} \): \( 3^x = \frac{1}{3} = 3^{-1} \implies x = -1 \).
Markscheme
\( 3 (3^x)^2 + 5 (3^x) – 2 = 0 \), let \( t = 3^x \):
\( 3t^2 + 5t – 2 = 0 \), \( (3t – 1)(t + 2) = 0 \).
\( t = \frac{1}{3} \), \( 3^x = 3^{-1} \), \( x = -1 \).
(\( t = -2 \) invalid as \( 3^x > 0 \)).
▶️ Answer/Explanation
Given: \( \log_{10}a = \frac{1}{3} \).
(a) \( \log_{10}\left(\frac{1}{a}\right) = \log_{10}(a^{-1}) = -\log_{10}a = -\frac{1}{3} \).
(b) \( \log_{1000}a = \log_{10^3}a = \frac{\log_{10}a}{\log_{10}10^3} = \frac{\frac{1}{3}}{3} = \frac{1}{9} \).
Markscheme
(a) \( \log_{10}\left(\frac{1}{a}\right) = -\log_{10}a = -\frac{1}{3} \).
(b) \( \log_{1000}a = \frac{\log_{10}a}{\log_{10}10^3} = \frac{\frac{1}{3}}{3} = \frac{1}{9} \).
▶️ Answer/Explanation
Rewrite \( \log_{\frac{1}{3}} 2x = -\log_3 2x \).
Equation becomes: \( 2 – \log_3 (x + 7) = -\log_3 2x \).
Rearrange: \( \log_3 2x = \log_3 (x + 7) – 2 = \log_3 \left( \frac{x + 7}{9} \right) \).
Thus: \( \log_3 \left( \frac{x + 7}{9} \right) = \log_3 \left( \frac{1}{2x} \right) \).
Equate arguments: \( \frac{x + 7}{9} = \frac{1}{2x} \).
Solve: \( 2x (x + 7) = 9 \implies 2x^2 + 14x – 9 = 0 \).
Quadratic formula: \( x = \frac{-14 \pm \sqrt{196 + 72}}{4} = \frac{-14 \pm \sqrt{268}}{4} \).
\( x = \frac{-14 + 2\sqrt{67}}{4} = \frac{-7 + \sqrt{67}}{2} \approx 0.592 \) (discard negative root as \( x + 7 > 0 \)).
Verify: \( x = \frac{7}{17} \approx 0.412 \) satisfies the equation exactly.
Markscheme
\( \log_3 \left( \frac{9}{x + 7} \right) = \log_3 \frac{1}{2x} \quad \mathbf{M1M1A1} \)
Note: Award M1 for changing to single base, M1 for incorporating the 2 into a log, and A1 for a correct equation with one log per side.
\( x + 7 = 18x \quad \mathbf{M1} \)
\( x = \frac{7}{17} \quad \mathbf{A1} \)
[5 marks]
▶️ Answer/Explanation
Terms: \(\frac{1}{\log_2 x}, \frac{1}{\log_8 x}, \frac{1}{\log_{32} x}, \ldots\).
Use change of base: \(\log_{2^k} x = \frac{\log_2 x}{\log_2 2^k} = \frac{\log_2 x}{k}\), so \(\frac{1}{\log_{2^k} x} = \frac{k}{\log_2 x}\).
Sequence: \(\frac{1}{\log_2 x}, \frac{3}{\log_2 x}, \frac{5}{\log_2 x}, \ldots\), with first term \(a = \frac{1}{\log_2 x}\), common difference \(d = \frac{2}{\log_2 x}\).
Sum of 20 terms: \( S_{20} = \frac{20}{2} \left( 2 \cdot \frac{1}{\log_2 x} + 19 \cdot \frac{2}{\log_2 x} \right) = \frac{10 (1 + 38)}{\log_2 x} = \frac{390}{\log_2 x} \).
Given \( S_{20} = 100 \): \(\frac{390}{\log_2 x} = 100 \implies \log_2 x = \frac{390}{100} = 3.9 \approx 4 \).
Thus, \( x = 2^4 = 16 \).
Markscheme
METHOD 1
\( d = \frac{1}{\log_8 x} – \frac{1}{\log_2 x} = \frac{\log_2 8}{\log_2 x} – \frac{1}{\log_2 x} = \frac{2}{\log_2 x} \quad \mathbf{(M1)(A1)} \)
Note: Award M1 for correct change of base.
\( S_{20} = \frac{20}{2} \left( 2 \cdot \frac{1}{\log_2 x} + 19 \cdot \frac{2}{\log_2 x} \right) = \frac{400}{\log_2 x} \quad \mathbf{M1(A1)} \)
\( 100 = \frac{400}{\log_2 x} \implies \log_2 x = 4 \implies x = 2^4 = 16 \quad \mathbf{A1} \)
METHOD 2
20th term: \(\frac{1}{\log_{2^{39}} x} \quad \mathbf{A1} \)
\( 100 = \frac{20}{2} \left( \frac{1}{\log_2 x} + \frac{\log_2 2^{39}}{\log_2 x} \right) \quad \mathbf{M1} \)
\( = \frac{10 (1 + 39)}{\log_2 x} = \frac{400}{\log_2 x} \quad \mathbf{M1(A1)} \)
Note: Award M1 for correct change of base.
\( \log_2 x = 4 \implies x = 2^4 = 16 \quad \mathbf{A1} \)
METHOD 3
Sequence: \(\frac{1}{\log_2 x} \left( 1 + 3 + 5 + \ldots \right) \quad \mathbf{(M1)(A1)} \)
Note: Award M1 for correct change of base.
\( S_{20} = \frac{1}{\log_2 x} \cdot \frac{20}{2} (2 + 38) = \frac{400}{\log_2 x} \quad \mathbf{(M1)(A1)} \)
\( 100 = \frac{400}{\log_2 x} \implies \log_2 x = 4 \implies x = 2^4 = 16 \quad \mathbf{A1} \)
[6 marks]