Question

Solve :
$y={\log }_{(x-4)}(x^2-11x+24)$

Answer 1

$y={log}_{(x-4)}(x^2-11x+24)$

$\Rightarrow y={log}_{(x-4)}(x^2-8x-3x+24)$

$\Rightarrow y={log}_{(x-4)}(x-8)(x-3)$

A. Solution will be possible only if

$x-4>0$ and $x-4\ne 1$

$\Rightarrow x>4$ and $x\ne 5⟶\left(1\right)$

And, $x^2-11x+24>0$

$\Rightarrow (x-8)(x-3)>0$

$\therefore$ $xϵ(-\infty ,3)\cup (8,\infty )⟶\left(2\right)$

$\therefore$ the acceptable values of ${}^{\prime }{x}^{\prime }$ for which $y={log}_{(x-4)}(x^2-11x+24)$ has a solution $10$ will be the intersection of the enequalities in $\left(1\right)$ & $\left(2\right)$.

$\therefore$ $xϵ(8,\infty )$