Consider the real-valued function $$f(x)=\frac{\log{(3x-7)}}{\sqrt{2x^{2}-7x+6}}$$ Find the domain of f(x).

The function $$f(x)=\frac{\log{(3x-7)}}{\sqrt{2x^{2}-7x+6}}$$ is only defined when both the numerator and the denominator of the function are defined are the denominator is not equal to zero.

The logarithm of the function is only defined for positive values :

Hence 3x-7 is greater than zero. Hence 

x > 7/3.

The value inside square root are defined for positive values. The value of the quadratic equation in the square root must be positive.

Hence $$2x^2-7x+6\ =\ 0$$ has the roots :

$$\frac{\left(7+\sqrt{\ 49-48}\right)}{4},\ \frac{\left(7-\sqrt{\ 49-48}\right)}{4}$$    : 2, 3/2

The quadratic equation is positive for :

$$\left(-\infty\ \frac{3}{2}\right)\ U\ \left(2,\ \infty\ \right)$$

Since in order to be a part of the domain the values of x must be greater than 7/3  and 7/3 is greater than 2 all values of x which are greater than 7/3 must be a part of the domain for x.