The function f(x) = x/logx increases on the interval

The correct option is (e, ∞).

To find the interval on which the function $f\left(x\right)=\frac{x}{logx}$ is increasing, we need to find the interval where its first derivative is positive, i.e., $f^{\text{'}}\left(x\right)>0$.

First, note that the function $f\left(x\right)=\frac{x}{logx}$ (where $logx$ denotes the natural logarithm, ${log}_{e}x$) is defined for $x>0$ and $x\ne 1$ (since $log1=0$).

Using the quotient rule of differentiation, ${\left(\frac{u}{v}\right)}^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$, let's differentiate $f\left(x\right)$ with respect to $x$:
$$f^{\text{'}}\left(x\right)=\frac{\frac{d}{dx}\left(x\right)·logx-x·\frac{d}{dx}(logx)}{{(logx)}^2}$$

Since $\frac{d}{dx}\left(x\right)=1$ and $\frac{d}{dx}(logx)=\frac{1}{x}$, we have:
$$f^{\text{'}}\left(x\right)=\frac{1·logx-x·\frac{1}{x}}{{(logx)}^2}$$

Simplifying the numerator:
$$f^{\text{'}}\left(x\right)=\frac{logx-1}{{(logx)}^2}$$

For the function $f\left(x\right)$ to be increasing, we require $f^{\text{'}}\left(x\right)>0$:
$$\frac{logx-1}{{(logx)}^2}>0$$

Since the denominator ${(logx)}^2$ is always positive for all $x$ in the domain of $f\left(x\right)$ (where $x\ne 1$), the sign of the derivative depends entirely on the numerator:

$$logx-1>0$$

Adding 1 to both sides:
$$logx>1$$

Taking the exponential of both sides (which is a strictly increasing function and preserves the inequality):
$$x>e^1$$
$$x>e$$

Thus, the function increases on the interval $(e,\infty )$.