2023 AISS AM S3 EOY Q8
2023 AISS AM S3 EOY Q8
9 marks
Others>EOY
- Solve the equation \({\log_e{(x^2 + 4x - 1)} = 2 + 2\log_e{(3x + 2)}}\)[5]
- Sketch the graphs of \({y = \log_e{x}}\) and \({y = e^{-x}}\) on the same axes. Use your graphs toexplain why \({\log_e{x} + e^{-x} > 0}\) for \({x > 1}\).[4]
Video Solution:
Video solution locked
A free account saves your progress; this video needs a course plan.
Solution:
Solution locked
680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs a course plan.
Create a free account to practise similar questions.
Answer:(a) \(x=9\)
(b) When \({x > 1}\), \({\log_2{x} > 0}\) \text{ and } \({e^{2x} > e^2 > 0}\).
Since both graphs/functions are always increasing,
\({\log_2{x} + e^{2x} > 0}\) for \({x > 1}\).
Need help? Join our A-Maths tuition classes.
Learn more