Q1. Knowledge: (a = 1.5 b = 1.5 = 3 marks) i) Find the value of each expression by using the definition of logarithm a) $\log_6 64$ b) $\log_3(\frac{1}{27})$ ii) Knowledge: ( = 2 marks) Identify the value of 'x' up to three decimal places. $3^{x+1} = 2^{2x-1}$
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Let $y = \log_6 64$. Then $6^y = 64$. Since $64 = 2^6$ and $6 = 2 \times 3$, we can't directly solve this equation. We can use the change of base formula: $\log_a b = \frac{\log_c b}{\log_c a}$. $\log_6 64 = \frac{\log 64}{\log 6} \approx \frac{1.806}{0.778} Show more…
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