Question

Solve the following inequality.
${\log }_{x-3}(x-4)<2.$

Answer 1

${\log }_{x-3}(x-4)<2$

$\frac{\log (x-4)}{\log (x-3)}<2$

$\Rightarrow \log (x-4)<2\log (x-3)$

$\Rightarrow \log (x-4)<\log (x-3{)}^2$

Remove log from both sides

$\Rightarrow x-4<(x-3{)}^2$

$\Rightarrow x^2+9-6x>x-4$

$\Rightarrow x^2-87x+13>0$

$\Rightarrow x^2-2\left(\frac{7}{2}\right)x+{\left(\frac{7}{2}\right)}^2-{\left(\frac{7}{2}\right)}^2+13>0$

(completing the square)

$\Rightarrow \underset{}{{(x-\frac{7}{2})}^2}+\frac{3}{4}>0$

This will always be positive for every value of $x$

Solution $\forall xϵR$

Interval notation $xϵ(-\infty ,\infty )$.

But for the logarithmic function to be defined, $x$ must be greater than $4$.

$\therefore$ Interval will be $xϵ(4,\infty )$.