Inequality exercise asked in the Lebanese University entrance Engineer Exam in 2001

Exercise:

Solve the following inequality, $\ln \left( \frac{x+1}{5-x} \right)>\ln \left( 2x-3 \right)$

Solution: we know that $\ln u\left( x \right)$ is defined when $u\left( x \right)>0$

So $\ln \left( \frac{x+1}{5-x} \right)$ is defined when $\frac{x+1}{5-x}>0$

Observe that $u\left( x \right)=\frac{x+1}{5-x}$ is defined when $5-x\ne 0\Leftrightarrow x\ne 5$  (*)

Now we need to study the sign of $u\left( x \right)$ using sign Table

                  \[\begin{align}
  & \underline{\left. x\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,-\infty \,\,\,\,\,\,\,\,\,\,\,\,-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\
 & \underline{\left. x+1\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,+\,\,\,\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
 & \underline{\left. 5-x\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,+\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
 & \left. Q\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,+\,\,\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\, \\
\end{align}\]

So $u\left( x \right)>0$ when $-1<x<5$  (**)

Also $\ln \left( 2x-3 \right)$ is defined when $2x-3>0\Leftrightarrow x>\frac{3}{2}=1.5$

Put $f\left( x \right)=\ln \left( \frac{x+1}{5-x} \right)-\ln \left( 2x-3 \right)$ is defined when $\frac{2}{3}<x<5$

Now $f\left( x \right)>0\Leftrightarrow \ln \left( \frac{x+1}{5-x} \right)-\ln \left( 2x-3 \right)>0\Leftrightarrow \ln \left( \frac{x+1}{5-x} \right)>\ln \left( 2x-3 \right)$

We have $\frac{x+1}{5-x}>\frac{2x-3}{1}\Leftrightarrow \frac{x+1-\left( 2x-3 \right)\left( 5-x \right)}{5-x}>0\Leftrightarrow \frac{-2{{x}^{2}}+12x-16}{5-x}>0$

$-2{{x}^{2}}+12x-16=0\Leftrightarrow -2\left( {{x}^{2}}-6x+8 \right)=0\Leftrightarrow {{x}^{2}}-6x+9-9+8=0\Leftrightarrow {{\left( x-3 \right)}^{2}}-1=0$

$\Leftrightarrow x-3=\pm 1\Leftrightarrow x=3\pm 1$ so $x=4\,or\,2$

By the sign table

         $\begin{align}
  & \underline{\left. x\,\,\,\,\,\,\,\,\, \right|\frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\
 & \underline{\left. E\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\, \\
 & \underline{\left. 5-x\, \right|\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
 & \left. Q\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,+\,\left. {} \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,- \\
\end{align}$

So the accepted solution is  $\frac{3}{2}<x<2\,\,\,\,\,or\,\,\,\,4<x<5$