another russian log problem

We would use here the fact that $\Large (a^m)^n=(a^n)^m=a^{mn}$ and $\Large a^{log_am}=m$ . Given that ---->

$\Large 4^{\log_{16}(4x^2-12x+9)}=(\frac{\sqrt{3}}{3})^{\log_{\frac{1}{3}}7 \centerdot \log_7(x^2+2x+1)}$

On squaring both sides ---->

$\Large (4^{\log_{16}(4x^2-12x+9)})^2=((\frac{\sqrt{3}}{3})^{\log_{\frac{1}{3}}7 \centerdot \log_7(x^2+2x+1)})^2$

$\Large \implies (4^2)^{\log_{16}(4x^2-12x+9)}=((\frac{\sqrt{3}}{3})^2)^{\log_{\frac{1}{3}}7 \centerdot \log_7(x^2+2x+1)}$

$\Large \implies 16^{\log_{16}(4x^2-12x+9)}=(\frac{3}{9})^{\log_{\frac{1}{3}}7 \centerdot \log_7(x^2+2x+1)}$

$\Large \implies 16^{\log_{16}(4x^2-12x+9)}=(\frac{1}{3})^{\log_{\frac{1}{3}}7 \centerdot \log_7(x^2+2x+1)}$

$\Large \implies 4x^2-12x+9=((\frac{1}{3})^{\log_{\frac{1}{3}}7})^{\log_7(x^2+2x+1)}$

$\Large \implies 4x^2-12x+9=7^{\log_7(x^2+2x+1)}$

$\implies 4x^2-12x+9=x^2+2x+1$

$\implies 3x^2-14x+8=0$

$\implies (x-4)(3x-2)=0$

$\implies \boxed{x=\{4,\frac{2}{3}\}}$

Sum of the solutions $=4+\frac{2}{3}=\frac{14}{3} \approx 4.67$

Since $4.67$ is closest to $5$ , so $5$ is our required answer.