Find a point on the x-axis,

Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

Accordingly, $\sqrt{(7-a)^{2}+(6-0)^{2}}=\sqrt{(3-a)^{2}+(4-0)^{2}}$

$\Rightarrow \sqrt{49+a^{2}-14 a+36}=\sqrt{9+a^{2}-6 a+16}$

$\Rightarrow \sqrt{a^{2}-14 a+85}=\sqrt{a^{2}-6 a+25}$

On squaring both sides, we obtain

$a^{2}-14 a+85=a^{2}-6 a+25$

$\Rightarrow-14 a+6 a=25-85$

$\Rightarrow-8 a=-60$

$\Rightarrow a=\frac{60}{8}=\frac{15}{2}$

Thus, the required point on the $x$-axis is $\left(\frac{15}{2}, 0\right)$.