Solve the equation ${\log_e{(x^2 + 4x - 1)} = 2 + 2\log_e{(3x + 2)}}$[5] Sketch the graphs of \({y

Question

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 to explain why ${\log_e{x} + e^{-x} > 0}$ for ${x > 1}$.[4]

Accepted Answer

(a) $x=9$ (b) When ${x > 1}$, ${\log_2{x} > 0}$ 	ext{ and } ${e^{2x} > e^2 > 0}$. Since both graphs/functions are always increasing, ${\log_2{x} + e^{2x} > 0}$ for ${x > 1}$.

2023 AISS AM S3 EOY Q8

2023 AISS AM S3 EOY Q8

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