Using the laws of logarithms, solve the equation
\[ \begin{aligned} 2\log_{2}(x - 3) - \log_{2}(6-x) &= 2 \end{aligned} \]
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SOLUTION:
\[ \begin{aligned} 2\log_{2}(x - 3) - \log_{2}(6-x) &= 2 \end{aligned} \tag{1} \] \[ \\ \\ \] \[ \begin{aligned} \;\;\Rightarrow\;\; \log_{2}\!\left(\frac{(x - 3)^{2}}{(6-x)}\right) &= 2 \\ \\ \frac{(x - 3)^{2}}{(6-x)} &= 2^{2} \\ \\ (x - 3)^{2} &= 4(6-x) \\ \\ x^{2} - 6x + 9 &= 24 - 4x \\ \\ x^{2} - 2x - 15 &= 0 \\ \\ \;\;\Rightarrow\;\; (x + 3)(x - 5) &= 0 \\ \\ \Rightarrow x &= -3,\; 5 \\ \\ \end{aligned} \] A logarithm is only defined for positive arguments: \[ \log_{b}(f(x)) \text{ exists only if } f(x) > 0 \] \[ \ \] Considering Terms 1 and 2 of Equation 1: \[ \begin{aligned} &\text{Term 1:} \;\; 2\log_{2}(x - 3) &\Rightarrow x - 3 > 0 &\Rightarrow x > 3 \\ \\ &\text{Term 2:} \;\; \log_{2}(6 - x) &\Rightarrow 6 - x > 0 &\Rightarrow x < 6 \end{aligned} \] \[ \ \] \[ \Rightarrow 3 < x < 6 \] \[ \\ \] \[ \therefore \underline{\underline{x = 5}} \]