\( \begin{array}{lc}\text { B-2 } \\ \text { 1. } & \log _{2}(x-2)(x-3)=1 \\ \text { 2. } & \log _{2}(4-x)=7 \\ \text { 3. } & \log _{5}(5-x)=\log _{5} 3 \\ \text { 4. } & \log _{5} x^{2}=\log _{5}(x+7) \\ \text { 5. } & \log _{\frac{1}{3}}(2 x-6)<\log _{\frac{1}{3}} x \\ \text { 6. } \quad \log _{\frac{1}{3}}(x-1) \geq-2\end{array} \)
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Let's dive right into solving these logarithmic equations and inequalities with some flair! For the first one, \( \log_2(x-2)(x-3) = 1 \): Here, we can convert the logarithmic equation into its exponential form. \( (x-2)(x-3) = 2^1 \), simplifying this leads to \( (x-2)(x-3) = 2 \). From the quadratic equation you derive, solving gives you the values for \( x \). Next, for \( \log_2(4-x) = 7 \): Transforming this one too gives us \( 4 - x = 2^7 \). Solving for \( x \) from \( 4 - x = 128 \) leads to \( x = -124 \). Just remember, sometimes you'll get those unusual answers! For \( \log_5(5-x) = \log_5 3 \): When you have equal logs, you set the insides equal: \( 5 - x = 3 \), leading to \( x = 2 \). Quick and easy! Now to \( \log_5 x^2 = \log_5(x+7) \): This means \( x^2 = x + 7 \), reorganizing it gives the quadratic \( x^2 - x - 7 = 0 \). Use the quadratic formula here to find \( x \). On to \( \log_{\frac{1}{3}}(2x-6) < \log_{\frac{1}{3}}x \): With the base being less than one, this flips the inequality: \( 2x - 6 > x \). Solve for \( x \) giving you \( x > 6 \). Finally, \( \log_{\frac{1}{3}}(x-1) \geq -2 \): Changing this also flips the inequality: \( x - 1 \leq \left(\frac{1}{3}\right)^{-2} = 9 \). So, \( x \leq 10 \), and don’t forget \( x > 1 \) from the log's domain. With logarithmic equations, it's all about understanding their properties and how to manipulate them, so keep practicing!