The number of integral solutions x of _(x+(7)/(2))((x-7)/(2x-3))^2 0 is :

Question

The number of integral solutions $x$ of ${\log }_{(x+\frac{7}{2})}{(\frac{x-7}{2x-3})}^{2}\ge 0$ is :

Accepted Answer

${\log }_{(x+\frac{7}{2})}{(\frac{x-7}{2x-3})}^{2}\ge 0$ Domain : $\begin{gathered} x+\frac{7}{2}>0 \ x>\frac{-7}{2} \ x+\frac{7}{2} e 1 \ x e \frac{-5}{2} \ \frac{x-7}{2x-3} e 0 \ x e 7 \ x e \frac{3}{2} \end{gathered}$ $x\in (\frac{-7}{2},\infty)-{-\frac{5}{2},\frac{3}{2},7}$ Now ${\log }_{a}b\ge 0$ if $a>1$ and $b\ge 1$ $\begin{gathered} a\in (0,1)b\in (0,1) \end{gathered}$ Case I : $x+\frac{7}{2}>1{(\frac{x-7}{2x-3})}^{2}\ge 1$ $ herefore $ $x>-\frac{5}{2}$ and $\begin{gathered} (2x-3{)}^{2}-(x-7{)}^{2}\le 0 \ (2x-3+x-7)(2x-3-x+7)\le 0 \ (3x-10)(x+4)\le 0 \end{gathered}$ $x\in [-4,\frac{10}{3}]$ $x\in (\frac{-5}{2},\frac{10}{3}]$ Case II : $x+\frac{7}{2}\in (0,1){(\frac{x-7}{2x-3})}^{2}\in (0,1)$ $\begin{gathered} herefore 0

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