Question

If ${\log }_{x-7}(x-4)$ is defined, then

• A. $x\in (8,\infty )$
• B. $x\in (4,7)$
• C. $x\in (4,\infty )-\left\{8\right\}$
• D. $x\in (7,\infty )-\left\{8\right\}$

Answer 1

The correct option is D $x\in (7,\infty )-\left\{8\right\}$

${\log }_{x-7}(x-4)$
Base $(x-7)>0$ and $x-7\ne 1$
$\Rightarrow x>7$ and $x\ne 8...\left(1\right)$

and $(x-4)>0$
$\Rightarrow x>4...\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$x\in (7,\infty )-\left\{8\right\}$