log _{left(x+frac{7}{2}right)}left(frac{x-7}{ | vedclass

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Question

$\log _{x+\frac{7}{2}}\left(\frac{x-7}{2 x-3}\right)^2 \geq 0$

Feasible region : $x+\frac{7}{2}>0 \Rightarrow x > -\frac{7}{2}$

And $x+\fr

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$\log _{x+\frac{7}{2}}\left(\frac{x-7}{2 x-3} ight)^2 \geq 0$ Feasible region : $x+\frac{7}{2}>0 \Rightarrow x > -\frac{7}{2}$ And $x+\frac{7}{2} eq 1 \Rightarrow x eq-\frac{5}{2}$ And $\frac{x-7}{2 x-3} eq 0 \quad$ and $2 x-3 eq 0$ $\Downarrow$ $x eq 7$ $x eq \frac{3}{2}$ Taking intersection : $x \in\left(\frac{-7}{2}, \infty ight)-\left\{-\frac{5}{2}, \frac{3}{2}, 7 ight\}$ Now $\log _a b \geq 0$ if $a > 1$ and $b \geq 1$ Or $a \in(0,1)$ and $b \in(0,1)$ $ ext { C-I; } \quad x+\frac{7}{2}>1 ext { and }\left(\frac{x-7}{2 x-3} ight)^2 \geq 1$ $x > -\frac{5}{2} \quad(2 x-3)^2-(x-7)^2 \leq 0$ $(2 x-3+n-7)(2 x-3-x+7) \leq 0$ $(3 x-10)(x+4) \leq 0$ $\quad x \in\left[-4, \frac{10}{3} ight]$ $ ext { Intersection : } x \in\left(\frac{-5}{2}, \frac{10}{3} ight]$ $ ext { C-II } x+\frac{7}{2} \in(0,1) ext { and }\left(\frac{x-7}{2 x-3} ight)^2 \in(0,1)$ $0 < x+\frac{7}{2} < 1 \quad \quad\left(\frac{x-7}{2 x-3} ight)^2 < 1$ $-\frac{7}{2} < x < \frac{-5}{2} \quad(x-7)^2<(2 x-3)^2$ $x \in(-\infty,-4) \cup\left(\frac{10}{3}, \infty ight)$ No common values of $x$. Hence intersection with feasible region We get $x \in\left(\frac{-5}{2}, \frac{10}{3} ight]-\left\{\frac{3}{2} ight\}$ Integral value of $x$ are $\{-2,-1,0,1,2,3\}$ No. of integral values $=6$

$\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^2 \geq 0$ નાં પૂર્ણાક ઉકેલો $x$ ની સંખ્યા $..........$ છે.

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