Solve the following equation for $x$ $\log _{(2 x+3)}\left(6 x^{2}+23 x+21\right)=4-\log _{3 x+7}\le

Question

Solve the following equation for $x$ $\log _{(2 x+3)}\left(6 x^{2}+23 x+21\right)=4-\log _{3 x+7}\left(4 x^{2}+12 x+9\right)$.

Professor PC Thomas Classes · Answer

We have, $2 x+3>0,2 x+3 
eq 1$\ $3 x+7>0,3 x+7 
eq 1$\ The given equation can be written as\ $\log _{(2 x+3)}(2 x+3)(3 x+7)$\ $=4-\log _{(3 x+7)}(2 x+3)^{2}$\ $\Rightarrow \quad \log _{2 x+3}(2 x+3)+\log _{(2 x+3)}(3 x+7)$\ $=4-2 \log _{(3 x+7)}(2 x+3)$ $$\begin{equation*}\Rightarrow \frac{\log (3 x+7)}{\log (2 x+3)}+\frac{2 \log (2 x+3)}{\log (3 x+7)}-3=0 	ag{1} \end{equation*}$$ Put $\frac{\log (2 x+3)}{\log (3 x+7)}=y$\ Then (1) can be written as $2 y+\frac{1}{y}-3=0$\ $\Rightarrow \quad 2 y^{2}-3 y+1=0$\ $\Rightarrow \quad(2 y-1)(y-1)=0$\ $\Rightarrow y=\frac{1}{2}$ (or) $y=1$\ When $y=\frac{1}{2}, \frac{\log (2 x+3)}{\log (3 x+7)}=\frac{1}{2}$\ $\Rightarrow \quad 2 \log (2 x+3)=\log (3 x+7)$\ $\Rightarrow \quad(2 x+3)^{2}=3 x+7$\ $\Rightarrow \quad 4 x^{2}+9+12 x=3 x+7$\ $\Rightarrow \quad 4 x^{2}+12 x-3 x+9-7=0$\ $\Rightarrow \quad 4 x^{2}+9 x+2=0 \Rightarrow(4 x+1)(x+2)=0$\ $\Rightarrow x=\frac{-1}{4}$ (or) $x=-2$\ When $y=1, \frac{\log (2 x+3)}{\log (3 x+7)}=1$\ $\Rightarrow \quad \log (2 x+3)=\log (3 x+7)$\ $\Rightarrow \quad 2 x+3=3 x+7 \Rightarrow x=-4$ Since $x>\frac{-3}{2}, x>\frac{-7}{3}, x 
eq-1, x 
eq-2$\ $	herefore x=\frac{-1}{4}$ is only solution of the given equation.\