The number of solutions of the equation [ _(x + 1)(2x^2 + 7x + 5) + _(2x + 5)(x + 1)^2 - 4…

Question

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

Answer: 1

Accepted Answer

$2x^2+7x+5=(2x+5)(x+1)$, so $\log_{x+1}(2x+5)+1+2\log_{2x+5}(x+1)-4=0 \implies \log_{x+1}(2x+5)+\dfrac2{\log_{x+1}(2x+5)}=3$.

Let $t=\log_{x+1}(2x+5)$: $t+\dfrac2t=3 \implies t^2-3t+2=0 \implies t=1,2$.

$t=1\Rightarrow 2x+5=x+1\Rightarrow x=-4$ (rejected, $x>0$). $t=2\Rightarrow2x+5=(x+1)^2\Rightarrow x^2=4\Rightarrow x=\pm2$, and $x=-2$ rejected. So $x=2$ is the only solution --- 1.

Solutions · 1