Question

$lo{g}_{(x-2)}(2x-3)>lo{g}_{(x-2)}(24-6x)$
The solution set of the above inequality has integral values of x ___

Answer 1

$\frac{27}{8}<x<4$ and $2<x<3$
We must have $2x-3>0,24-6x>0$
or $x>\frac{3}{2}$ and $x<4$
Also base $x-2>0\Rightarrow x>2$
Hence we must have $x>2$ and $x<4\dots \dots \left(1\right)$
Now if base x-2>1 i.e. x > 3, then we must have
$2x-3>24-6x$
or $8x>27\therefore x>\frac{27}{8}\dots \dots \left(2\right)$
Hence from (1) and (2), we have
$\frac{27}{8}<x<4$
However if $0<x-2<1$ or $2<x<3$ then we
must have $2x-3<24-6x$ or $8x<27\therefore x<\frac{27}{8}$
In this case we have $2<x<3\dots \dots \left(B\right)$
Both (A) and (B) give the required solution