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Q. Solve ${\log }_3\left|3-4x\right|>2$

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$$\begin{gathered} To Evaluate\hspace{0.17em}: {\log }_3\left|3-4x\right|>2 \\ Solution : {\log }_3\left|3-4x\right|>2 \\ \Rightarrow \left|3-4x\right|>3^2 \\ \Rightarrow \left|3-4x\right|>9 \\ here we have two cases \\ case \left(1\right) x<\frac{3}{4} \\ so, putting a value less than \frac{3}{4} to check whether modulus opens with positive or negative sign \\ so, putting x=0 \\ \left|3-4\times 0\right|=\left|3-0\right|=\left|3\right|=3 \\ So, modulus will opem with positive sign \\ \Rightarrow +(3-4x)>9 \\ \Rightarrow 3-4x>9 \\ \Rightarrow 3-9>4x \\ \Rightarrow -6>4x \\ \Rightarrow x<\frac{-6}{4} \\ \Rightarrow x<\frac{-3}{2} \\ Since we are checking in x<\frac{3}{4} and x<\frac{-3}{2} lies in the range. \\ So, overall x<\frac{-3}{2} \\ \Rightarrow x\in (-\infty , -\frac{3}{2}) \\ case \left(2\right) x\ge \frac{3}{4} \\ so, putting a value greater than \frac{3}{4} to check whether modulus opens with positive or negative sign \\ so, putting x=1 \\ \left|3-4\times 1\right|=\left|3-4\right|=\left|-1\right|=-(-1)=1 \\ So, modulus will opem with negative sign \\ \Rightarrow -(3-4x)>9 \\ \Rightarrow -3+4x>9 \\ \Rightarrow 4x>9+3 \\ \Rightarrow x>\frac{12}{4} \\ \Rightarrow x>3 \\ Since we are checking in x>\frac{3}{4} and x>3 lies in the range. \\ So, overall x>3 \\ x\in (3, \infty ) \\ So, complete solution will be the union of both the cases \\ \Rightarrow x\in (-\infty , -\frac{3}{2})\cup (3, \infty ) \end{gathered}$$
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