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

We have,

$2^x+2^{x-1}+2^{x-2}=7^x+7^{x-1}+7^{x-2}$

$⇒2^x(1+\frac{1}{2}+\frac{1}{4})=7^x(1+\frac{1}{7}+\frac{1}{49})$

$⇒2^x×\frac{7}{4}=7^x×\frac{57}{49}$

$⇒2^{x-2}=7^{x-2}×\frac{57}{7}$

$⇒(\frac{7}{2})^{x-2}=\frac{7}{57}$

$⇒x-2=\log_{(7/2)}(\frac{7}{57})⇒x=2+\log_{7/2}(\frac{7}{57})$

$y_1=2^x+2^{x-1}+2^{x-2}$   $y_2=7^x+7^{x-1}+7^{x-2}$

$y_1=2^x(1+\frac{1}{2}+\frac{1}{4})$   $y_2=7^x(1+\frac{1}{7}+\frac{1}{49})$

Hence, the given equation has one root.