Solve the equation $\log_2 (x^2 + 4x - 1) = 2 + 2\log_4 (3x + 2)$.[5] Sketch the graphs of \(y = \

Question

Solve the equation $\log_2 (x^2 + 4x - 1) = 2 + 2\log_4 (3x + 2)$.[5] Sketch the graphs of $y = \log_2 x$ and $y = e^{2x}$ on the same axes. Use your graphs to explain why $\log_2 x + e^{2x} > 0$ for $x > 1$.[4]

Accepted Answer

(a) $x=9$ (b) Both $\log_2x$ and $e^{2x}$ are positive for $x>1$, so $\log_2x+e^{2x}>0$

2023 Ahmad Ibrahim Sec AM S3 EOY Q8

2023 Ahmad Ibrahim Sec AM S3 EOY Q8

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