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Question

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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

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

The number of solutions of the equation $\log _{(x+1)}\left(2 x^{2}+7 x+5\right)+\log _{(2 x+5)}(x+1)^{2}-4=0, x\,>\,0$, is $....$

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