Question

Solve the following inequality:
$\frac{1+{\log }_{x+1}(x-3)}{{\log }_{x+1}3}<{\log }_3(2x-3)$

Answer 1

$\frac{1+{\log }_{x+1}(x-3)}{{\log }_{(x+1)}\left(3\right)}<{\log }_3(2x-3)$

Here,

$x+1>0x-3>02x-3>0$

$x>-1x>3x>1.5$

Hence, taking intersection $x>3\Rightarrow xϵ(3,\infty )$

For $xϵ(3,\infty )$

$1+{\log }_{x+1}(x-3)<1+{\log }_3(2x-3){\log }_{(x+1)}3$

$1+{\log }_{(x+1)}(x-3)<{\log }_{x+1}(2x-3)$

${\log }_{(x+1)}(x+1)+{\log }_{(x+1)}(x-3)<{\log }_{(x+1)}(2x-3)$

$(x+1)(x-3)<(2x-3)$

$x^2-2x-3<2x-3$

$x^2-4x<0\Rightarrow x(x-4)<0$

$xϵ(0,4)$

Taking intersection we get $xϵ(3,4)$