If  $5^x+{\left(2\sqrt{3}\right)}^{2x}={13}^x,$  then the solution set for $x$ is

$5^x+{\left(2\sqrt{3}\right)}^{2x}=5^x+{12}^x\ge {13}^x$

$\Rightarrow$  ${\left(\frac{5}{13}\right)}^x+{\left(\frac{12}{13}\right)}^x\ge 1$

Equality holds for $x=2$

As   $\frac{5}{13}<1,{\left(\frac{5}{13}\right)}^x>{\left(\frac{5}{13}\right)}^2$    for  $x<2$

$\frac{12}{13}<1,$   ${\left(\frac{12}{13}\right)}^x>{\left(\frac{12}{13}\right)}^2$  for $x<2$

$\therefore$  For $x<2,$

${\left(\frac{5}{13}\right)}^x+{\left(\frac{12}{13}\right)}^x>{\left(\frac{5}{13}\right)}^2+{\left(\frac{12}{13}\right)}^2=1$

Hence required solution is $\left(-\infty ,2\right]$