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7. (i) Use logarithms to solve the equation $8^{2x+1} = 24$, giving your answer to 3 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 2
Question 8
7. (i) Use logarithms to solve the equation $8^{2x+1} = 24$, giving your answer to 3 decimal places. (ii) Find the values of $y$ such that \[ \log_{2}(11y - 3) - \l... show full transcript
Worked Solution & Example Answer:7. (i) Use logarithms to solve the equation $8^{2x+1} = 24$, giving your answer to 3 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 2
Step 1
(i) Use logarithms to solve the equation $8^{2x+1} = 24$
Answer
To solve the equation, we can take logarithms on both sides:
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Apply logarithms: [ \log(8^{2x+1}) = \log(24) ]
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Using the power rule of logarithms: [ (2x+1) \log(8) = \log(24) ]
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Isolate : [ 2x + 1 = \frac{\log(24)}{\log(8)} ] Now calculating the right-hand side: [ \log(8) = 3 \log(2), \quad \log(24) \approx 1.3802 ]
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Thus we have: [ 2x + 1 = \frac{1.3802}{3 \log(2)} \approx 0.5783 ]
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This results in: [ 2x = 0.5783 - 1 \approx -0.4217 ]
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Finally, solving for : [ x \approx \frac{-0.4217}{2} \approx -0.2108 ]
Rounding to three decimal places, we get: [ x \approx -0.211 ]
Step 2
(ii) Find the values of $y$ such that \log_{2}(11y - 3) - \log_{2} 3 - 2 \log_{2} y = 1
Answer
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Begin by using the properties of logarithms: [ \log_{2}(11y - 3) - \log_{2} 3 = 1 + 2 \log_{2} y ]
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Applying the properties gives: [ \log_{2}\left(\frac{11y - 3}{3}\right) = 1 + \log_{2}(y^2) ]
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Exponentiate both sides: [ \frac{11y - 3}{3} = 2(y^2) ]
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Rearranging this leads to: [ 11y - 3 = 6y^2 ]
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Rearranging to form a quadratic equation gives: [ 6y^2 - 11y + 3 = 0 ]
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Using the quadratic formula: [ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ] Where , , : [ y = \frac{11 \pm \sqrt{(-11)^2 - 4 \times 6 \times 3}}{2 \times 6} = \frac{11 \pm \sqrt{121 - 72}}{12} = \frac{11 \pm \sqrt{49}}{12} = \frac{11 \pm 7}{12} ]
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This yields: [ y = \frac{18}{12} = 1.5 \quad \text{and} \quad y = \frac{4}{12} = \frac{1}{3} ]
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Therefore, considering the condition , only: [ y = 1.5 \text{ is acceptable.} ]