jmc

The square root $\sqrt{3^x-7^x}$ is defined only when $3^x\ge 7^x.$ This is equivalent to $\frac{7^x}{3^x}\le 1,$ or $${\left(\frac{7}{3}\right)}^x\le 1.$$This inequality is satisfied exactly when $x\le 0.$ Thus, the domain of the function is $\boxed{(-\infty ,0]}.$