Logarithms and Properties
4.0 Logarithmic Inequalities
4.0 Logarithmic Inequalities
Consider two logarithmic functions, ${\log _a}p$ & ${\log _a}q$
Let, $$\begin{equation} \begin{aligned} {\log _a}p = x\quad or\quad {a^x} = p \\ {\log _a}q = y\quad or\quad {a^y} = q \\\end{aligned} \end{equation} $$
If, $x>y$
So, we will have different case depending on the value of the base to the logarithmic functions.
Case 1: When $a>1$
For $x>y$,
Then, ${a^x} > {a^y}$
Therefore, $p>q$
Case 2: When $0<a<1$
For $x>y$,
Then, ${a^x} < {a^y}$
Therefore, $p<q$
Also remember
- If $a>1$, $p>1$, then ${\log _a}p > 0$
- If $0<a<1$, $p>1$, then ${\log _a}p < 0$
- If $a>1$, $0<p<1$, then ${\log _a}p < 0$
- If $p>a>1$, then ${\log _a}p > 0$
- If $a>p>1$, then ${\log _a}p < 0$
- If $0<a<p<1$, then $0<{\log _a}p > 1$
- If $0<p<a<1$, then ${\log _a}p > 1$
Question 9. Solve for $x$, ${\log _{\frac{1}{2}}}(x - 2) > 2$
Solution: For logarithm to be defined, $$\begin{equation} \begin{aligned} (x - 2) > 0 \\ x > 2\quad ...(i) \\\end{aligned} \end{equation} $$So, $$\begin{equation} \begin{aligned} {\log _{\frac{1}{2}}}(x - 2) > 2 \\ {\log _{\frac{1}{2}}}(x - 2) > {\log _{\frac{1}{2}}}{\left( {\frac{1}{2}} \right)^2} \\\end{aligned} \end{equation} $$ Since, the base $\left( {a = \frac{1}{2}} \right)$ is less than 1, so the equality reverses while removing solving log. $$\begin{equation} \begin{aligned} (x - 2) < {\left( {\frac{1}{2}} \right)^2} \\ (x - 2) < \frac{1}{4} \\ 4x - 8 < 1 \\ 4x < 9 \\ x < \frac{9}{4}\quad ...(ii) \\\end{aligned} \end{equation} $$ From equation $(i)$ & $(ii)$ we get, $$x:\left( {2,\frac{9}{4}} \right)$$
