Question

The solution set of domain ${\log }_{\mathrm{e}}\left(\frac{x-2}{x-3}\right)$ is

• A. $\left(2,\infty \right)$
• B. $\left(-\infty ,2\right)$
• C. $\left(-\infty ,\infty \right)$
• D. $\left(4,\infty \right)$

Answer 1

Answer: (B), (D)

$\left(4,\infty \right)$

Explanation for correct option:

We know that ${\log }_{\mathrm{e}}\frac{\mathrm{a}}{\mathrm{b}}$ is defined only when $\mathrm{a}>0,\mathrm{b}>0$ or $\mathrm{a}<0,\mathrm{b}<0$

So, ${\log }_{\mathrm{e}}\left(\frac{x-2}{x-3}\right)$ is defined when $\mathrm{x}-2>0,\mathrm{x}-3>0$

$\Rightarrow$$\mathrm{x}>2,\mathrm{x}>3$

$\Rightarrow$$\mathrm{x}>3$ $...\left(\mathrm{i}\right)$

Or, ${\log }_{\mathrm{e}}\left(\frac{x-2}{x-3}\right)$ is defined when $\mathrm{x}-2<0,\mathrm{x}-3<0$

$\Rightarrow$$\mathrm{x}<2,\mathrm{x}<3$

$\Rightarrow$$\mathrm{x}<2$ $...\left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get; $\mathrm{x}\in \left(-\infty ,2\right)\cup \left(3,\infty \right)$

Thus option(B) and option (D) is correct

Hence, the correct option is option(B) and option (D).