# Logarithmic Inequalities

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}