Algebra exercise about log function of base

Exercise:

Solve in $\mathbb{R},$ ${{\log }_{3x+4}}\left( 4{{x}^{2}}+4x+1 \right)+{{\log }_{2x+1}}\left( 6{{x}^{2}}+11x+4 \right)=4$

Solution: we know that , ${{\log }_{\varphi \left( x \right)}}u\left( x \right)$ is defined whenever $\varphi \left( x \right)>0$ and $u\left( x \right)>0$

Hence $3x+4>0\Leftrightarrow x>-\frac{4}{3}\,\,\,\,\And \,\,\,2x+1>0\Leftrightarrow x>-\frac{1}{2}$

${{\log }_{3x+4}}\left( 4{{x}^{2}}+4x+1 \right)=\frac{\ln \left( 4{{x}^{2}}+4x+1 \right)}{\ln \left( 3x+4 \right)}=\frac{\ln {{\left( 2x+1 \right)}^{2}}}{\ln \left( 3x+4 \right)}=\frac{2\ln \left( 2x+1 \right)}{\ln \left( 3x+4 \right)}$

${{\log }_{2x+1}}\left( 6{{x}^{2}}+11x+4 \right)=\frac{\ln \left( 6{{x}^{2}}+11x+4 \right)}{\ln \left( 2x+1 \right)}=\frac{\ln \left[ \left( 3x+4 \right)\left( 2x+1 \right) \right]}{\ln \left( 2x+1 \right)}$

Hence $\frac{2\ln \left( 2x+1 \right)}{\ln \left( 3x+4 \right)}+\frac{\ln \left( 3x+4 \right)+\ln \left( 2x+1 \right)}{\ln \left( 2x+1 \right)}=4$

$\Rightarrow \frac{2\ln \left( 2x+1 \right)}{\ln \left( 3x+4 \right)}+\frac{\ln \left( 3x+4 \right)}{\ln \left( 2x+1 \right)}=4-1=3$

Let $w=\frac{\ln \left( 2x+1 \right)}{\ln \left( 3x+4 \right)}\Rightarrow {{w}^{-1}}=\frac{\ln \left( 3x+4 \right)}{\ln \left( 2x+1 \right)}$

Thus $2w+{{w}^{-1}}=3\Leftrightarrow 2{{w}^{2}}+1=3w\Leftrightarrow 2{{w}^{2}}-3w+1=0\Leftrightarrow \left( w-1 \right)\left( 2w-1 \right)=0$

So $w=1\,\And \,w=\frac{1}{2}$

But $\frac{\ln \left( 2x+1 \right)}{\ln \left( 3x+4 \right)}=1\Leftrightarrow \ln \left( 2x+1 \right)=\ln \left( 3x+4 \right)\Leftrightarrow 2x+1=3x+4\Leftrightarrow -x=3\Leftrightarrow x=-3$ rejected

And $\frac{\ln \left( 2x+1 \right)}{\ln \left( 3x+4 \right)}=\frac{1}{2}\Rightarrow \ln {{\left( 2x+1 \right)}^{2}}=\ln \left( 3x+4 \right)\Leftrightarrow 4{{x}^{2}}+4x+1=3x+4\Leftrightarrow 4{{x}^{2}}+x-3=0$

$\Leftrightarrow {{x}^{2}}+\frac{1}{4}x-3=\left( x+1 \right)\left( 4x-3 \right)=0$thus $x=-1$ rejected and $x=\frac{3}{4}$ is accepted .