4-2.Quadratic Equations and Inequations
hard

समीकरण

$\log _{(x+1)}\left(2 x^{2}+7 x+5\right)+\log _{(2 x+5)}(x+1)^{2}-4=0 \text {, }$

$x > 0$ के हलों की संख्या है .............. 

A

$2$

B

$4$

C

$6$

D

$1$

(JEE MAIN-2021)

Solution

$\log _{(x+1)}\left(2 x^{2}+7 x+5\right)+\log _{(2 x+5)}(x+1)^{2}-4=0$

$\log _{(x+1)}(2 x+5)(x+1)+2 \log _{(2 x+5)}(x+1)=4$

$\log _{(x+1)}(2 x+5)+1+2 \log _{(2 x+5)}(x+1)=4$

$\text { Put } \log _{(x+1)}(2 x+5)=t$

$t+\frac{2}{t}=3 \Rightarrow t^{2}-3 t+2=0$

$t=1,2$

$\log _{(x+1)}(2 x+5)=1 \quad \& \log _{(x+1)}(2 x+5)=2$

$x+1=2 x+3 \quad \& \quad 2 x+5=(x+1)^{2}$

$x=-4$ (rejected) $\quad x^{2}=4 \Rightarrow x=2,-2$ (rejected)

So, $x=2$

No. of solution $=1$

Standard 11
Mathematics

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