${(2x^2+3x+4)}^{10}=\sum _{r=0}^{20}{a}_r{x}^r$

Replace $x\to \frac{2}{x}, {(\frac{8}{x^2}+\frac{6}{x}+4)}^{10}=\sum _{r=0}^{20}{a}_r{\left(\frac{2}{x}\right)}^r$.
$\Rightarrow \frac{2^{10}}{x^{20}}{\left[2x^2+3x+4\right]}^{10}=\sum _{r=0}^{20} a_r{2}^r\left(\frac{1}{x^r}\right)$

$2^{10}\sum _{r=0}^{20}{a}_r{x}^r=\sum _{r=0}^{20}{a}_r{2}^r{x}^{20-r}$

Coefficient of  $x^7$ is $2^{10}{a}_7=a_{13}{2}^{13}\Rightarrow \frac{a_7}{a_{13}}=8$