Pls solve the question and answer with detail.:

4. The number of negative integral value of x satisfying the inequality $\log_{(x + \frac{5}{2})} {(\frac{x - 5}{2 x - 3})}^{2} < 0 is .$

\log_{(x + \frac{5}{2})} {(\frac{x - 5}{2 x - 3})}^{2} < 0 B a s e o f s h o u l d b e g r e a t e r t h a n z e r o a n d n o t e q u a l t o 1 x + \frac{5}{2} > 0 a n d x + \frac{5}{2} \neq 1 x > - \frac{5}{2} a n d x \neq - \frac{3}{2} C a s e 1 : x + \frac{5}{2} > 1 i . e . x > - \frac{3}{2} U \sin g \log a r i t h m i c i n e q u a l i t y f o r b a s e > 1 . \log_{a} b < c \Rightarrow 0 < b < a^{c} \log_{(x + \frac{5}{2})} {(\frac{x - 5}{2 x - 3})}^{2} < 0 0 < {(\frac{x - 5}{2 x - 3})}^{2} < {(x + \frac{5}{2})}^{0} 0 < {(\frac{x - 5}{2 x - 3})}^{2} < 1 0 < {(\frac{x - 5}{2 x - 3})}^{2} i s a l w a y s t r u e f o r x \neq 5 . A l s o d e n o m i n a t o r s h o u l d n o t b e z e r o s o 2 x - 3 \neq 0 i . e . x \neq \frac{3}{2} {(\frac{x - 5}{2 x - 3})}^{2} < 1 {(\frac{x - 5}{2 x - 3})}^{2} - 1 < 0 \frac{{(x - 5)}^{2} - {(2 x - 3)}^{2}}{{(2 x - 3)}^{2}} < 0 \frac{(x^{2} + 25 - 10 x) - (4 x^{2} + 9 - 12 x)}{{(2 x - 3)}^{2}} < 0 \frac{x^{2} + 25 - 10 x - 4 x^{2} - 9 + 12 x}{{(2 x - 3)}^{2}} < 0 \frac{- 3 x^{2} + 16 + 2 x}{{(2 x - 3)}^{2}} < 0 \frac{3 x^{2} - 2 x - 16}{{(2 x - 3)}^{2}} > 0 \frac{3 x^{2} + 6 x - 8 x - 16}{{(2 x - 3)}^{2}} > 0 \frac{3 x (x + 2) - 8 (x + 2)}{{(2 x - 3)}^{2}} > 0 \frac{(x + 2) (3 x - 8)}{{(2 x - 3)}^{2}} > 0 \frac{3 (x + 2) (x - \frac{8}{3})}{4 {(x - \frac{3}{2})}^{2}} > 0 \frac{(x - (- 2)) (x - \frac{8}{3})}{{(x - \frac{3}{2})}^{2}} > 0 D r a w i n g w a v y c u r v e

x \in (- \infty, - 2) \cup (\frac{8}{3}, \infty) B u t x > - \frac{3}{2} a n d x \neq 5, \frac{3}{2} x \in (\frac{8}{3}, 5) \cup (5, \infty)

C a s e 2 : 0 < x + \frac{5}{2} < 1 i . e . - \frac{5}{2} < x < - \frac{3}{2} U \sin g \log a r i t h m i c i n e q u a l i t y f o r b a s e < 1 . \log_{a} b < c \Rightarrow b > a^{c} \log_{(x + \frac{5}{2})} {(\frac{x - 5}{2 x - 3})}^{2} < 0 {(\frac{x - 5}{2 x - 3})}^{2} > {(x + \frac{5}{2})}^{0} {(\frac{x - 5}{2 x - 3})}^{2} > 1 {(\frac{x - 5}{2 x - 3})}^{2} - 1 > 0 \frac{{(x - 5)}^{2} - {(2 x - 3)}^{2}}{{(2 x - 3)}^{2}} > 0 \frac{(x^{2} + 25 - 10 x) - (4 x^{2} + 9 - 12 x)}{{(2 x - 3)}^{2}} > 0 \frac{x^{2} + 25 - 10 x - 4 x^{2} - 9 + 12 x}{{(2 x - 3)}^{2}} > 0 \frac{- 3 x^{2} + 16 + 2 x}{{(2 x - 3)}^{2}} > 0 \frac{3 x^{2} - 2 x - 16}{{(2 x - 3)}^{2}} < 0 \frac{3 x^{2} + 6 x - 8 x - 16}{{(2 x - 3)}^{2}} < 0 \frac{3 x (x + 2) - 8 (x + 2)}{{(2 x - 3)}^{2}} < 0 \frac{(x + 2) (3 x - 8)}{{(2 x - 3)}^{2}} < 0 \frac{3 (x + 2) (x - \frac{8}{3})}{4 {(x - \frac{3}{2})}^{2}} < 0 \frac{(x - (- 2)) (x - \frac{8}{3})}{{(x - \frac{3}{2})}^{2}} < 0 D r a w i n g w a v y c u r v e

x \in (- 2, \frac{3}{2}) \cup (\frac{3}{2}, \frac{8}{3}) B u t - \frac{5}{2} < x < - \frac{3}{2} x \in (- 2, - \frac{3}{2}) H e n c e c o m p l e t e s o l u t i o n s e t w i l l b e x \in (- 2, - \frac{3}{2}) \cup (\frac{8}{3}, 5) \cup (5, \infty)

Pls solve the question and answer with detail.: 4. The number of negative integral value of x satisfying the inequality log x + 5 2 x - 5 2 x - 3 2 < 0 is .