In solving logarithmic inequalities, it is important to understand the direction of the inequality changes if the base of the logarithms is less than 1.
$$\log_{2}{x} \lt \log_{2}{y}, \text{ then } x \lt y \\
\log_{0.5}{x} \lt \log_{0.5}{y}, \text{ then } x \gt y \\
$$
Also, the domain of the logarithm is positive.
$$\log_{10}{(x-2)}, \text{ then } x-2 \gt 0$$
Question 1
Solve \( \log_{3}{(x-3)} \gt \log_{3}{(x-1)}. \)
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\text{domain: } x-3 &\gt 0 \text{ or } x-1 \gt 0 \\
\text{that is } x &\gt 3 \color{green} \cdots (1) \\
x-3 &\gt x-1 \\
x &\gt 2 \color{green} \cdots (2) \\
\therefore x &\gt 3 &\color{green} \text{ by (1) and (2)}\\
\end{aligned} \\ \)
Question 2
Solve \( \log_{3}{(x-3)} \gt \log_{9}{(x-1)} \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\text{domain: } x-3 &\gt 0 \text{ or } x-1 \gt 0 \\
\text{that is } x &\gt 3 \color{green} \cdots (1) \\
\log_{3}{(x-3)} &\gt \frac{\log_{3}{(x-1)}}{\log_{3}{9}} &\color{green} \log_{a}{b} = \frac{\log_{c}{b}}{\log_{c}{a}} \\
\log_{3}{(x-3)} &\gt \frac{\log_{3}{(x-1)}}{\log_{3}{3^2}} \\
\log_{3}{(x-3)} &\gt \frac{\log_{3}{(x-1)}}{2\log_{3}{3}} \\
\log_{3}{(x-3)} &\gt \frac{\log_{3}{(x-1)}}{2} \\
2 \log_{3}{(x-3)} &\gt \log_{3}{(x-1)} \\
\log_{3}{(x-3)^2} &\gt \log_{3}{(x-1)} \\
(x-3)^2 &\gt x-1 \\
x^2-6x + 9 &\gt x-1 \\
x^2-7x + 10 &\gt 0 \\
(x-2)(x-5) &\gt 0 \\
x &\lt 2 \text{ or } x \gt 5 \color{green} \cdots (2) \\
\therefore x &\gt 5 &\color{green} \text{by (1) and (2) } \\
\end{aligned} \\ \)
Question 3
Solve \( \log_{0.5}{(x^2-19)}-\log_{0.5}{(x-5)} \lt \log_{0.5}{5} \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\text{domain: } x^2-19 &\gt 0 \text{ or } x-5 \gt 0 \\
\text{that is } x &\gt 5 \color{green} \cdots (1) \\
\log_{0.5}{(x^2-19)} &\lt \log_{0.5}{5} + \log_{0.5}{(x-5)} \\
\log_{0.5}{(x^2-19)} &\lt \log_{0.5}{5(x-5)} \\
x^2-19 &\gt 5(x-5) &\color{green} \text{direction changed as base } \lt 1 \\
x^2-19 &\gt 5x-25 \\
x^2-5x +6 &\gt 0 \\
(x-2)(x-3) &\gt 0 \\
x &\lt 2 \text{ or } x \gt 3 \color{green} \cdots (2) \\
\therefore x &\gt 5 &\color{green} \text{ by (1) and (2) } \\
\end{aligned} \\ \)
Question 4
Solve \( (\log_{3}{x})^2 \lt \log_{3}{x^4} \).
\( \begin{aligned} \displaystyle
(\log_{3}{x})^2 &\lt 4\log_{3}{x} \\
(\log_{3}{x})^2-4\log_{3}{x} &\lt 0 \\
\log_{3}{x}(\log_{3}{x}-4) &\lt 0 \\
0 &\lt \log_{3}{x} \lt 4 \\
3^0 &\lt x \lt 3^4 \\
\therefore 1 &\lt x \lt 81 \\
\end{aligned} \\ \)
Question 5
Solve \( x^{\log_{2}{x}} \lt 8 x^2 \).
\( \begin{aligned} \displaystyle
\log_{2}{x^{\log_{2}{x}}} &\lt \log_{2}{8x^2} \\
\log_{2}{x} \times \log_{2}{x} &\lt \log_{2}{8} + \log_{2}{x^2} \\
(\log_{2}{x})^2 &\lt \log_{2}{2^3} + 2\log_{2}{x} \\
(\log_{2}{x})^2 &\lt 3 + 2\log_{2}{x} \\
(\log_{2}{x})^2-2\log_{2}{x}-3 &\lt 0 \\
(\log_{2}{x} + 1)(\log_{2}{x}-3) &\lt 0 \\
-1 &\lt \log_{2}{x} \lt 3 \\
2^{-1} &\lt x \lt 2^3 \\
\therefore \frac{1}{2} &\lt x \lt 8 \\
\end{aligned} \)
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