Given that $\log_y 7 = 2 \log_g 4 + \frac{1}{2}$, find $y$ in terms of $a$ - AQA - A-Level Maths Pure - Question 7 - 2018 - Paper 3

Starting with the given equation: $\log_y 7 = 2 \log_g 4 + \frac{1}{2}$ We can write it as: $\log_y 7 = \log_g 4^2 + \log_g 4^{1/2}$ This can be simplified to: $\log_y 7 = \log_g \left( (4^2) \cdot (4^{1/2}) \right)$ Calculating, $\log_y 7 = \log_g \left( 49 \cdot 4 \right) = \log_g 196$ Using the change of base formula, we get: $\log_y 7 = \frac{\log_g 196}{\log_g y}$ Hence, $\log_g y = \frac{\log_g 196}{\log_g 7}$ So, $y = \log_g 196 \cdot \log_g 7^{-1} = \log_g 196 \cdot \frac{1}{\log_g 7}$ In terms of $a$, we can denote: $\log_g 196 = \log_g \left( 49 \cdot 4 \right) = \log_g 4^3 = 3 \log_g 4$ Thus: $y = 3 \log_g 4 / \log_g 7$