The number of roots of the equation $2^x+2^{x-1}+2^{x-2}=7^x+7^{x-1}+7^{x-2}$  is:

$2^x\left(1+\frac{1}{2}+\frac{1}{4}\right)=7^x\left(1+\frac{1}{7}+\frac{1}{7^2}\right)$

$\Rightarrow$       $2^x\left(\frac{7}{4}\right)=7^x\left(\frac{57}{49}\right)$

$\Rightarrow$        ${\left(\frac{7}{2}\right)}^{x-2}=\left(\frac{7}{57}\right)$

or           $\left(x-2\right)\log \left(\frac{7}{2}\right)=\log \left(\frac{7}{57}\right)$

$\therefore$            $x=2+\frac{\log \left(\frac{7}{57}\right)}{\log \left(\frac{7}{2}\right)}$