Let R+ be the set of all positive real numbers.

Given that $\mathrm{f}: \mathrm{R}+\rightarrow \mathrm{R}$ such that $\mathrm{f}(\mathrm{x})=\log _{\mathrm{e}} \mathrm{x}$

To find: (i) Range of $f$

Here, $f(x)=\log _{e} x$

We know that the range of a function is the set of images of elements in the domain.

$\therefore$ The image set of the domain of $\mathrm{f}=\mathrm{R}$

Hence, the range of f is the set of all real numbers.

To find: (ii) $\left\{x: x \in R^{+}\right.$and $\left.f(x)=-2\right\}$

We have, $f(x)=-2 \ldots$ (a)

And $f(x)=\log _{e} x \ldots(b)$

From eq. (a) and (b), we get

$\log _{e} x=-2$

Taking exponential both the sides, we get

$\Rightarrow e^{\log _{e} x}=e^{-2}$

$\left[\because\right.$ Inverse property $\cdot$ i. e $\left.b^{\log _{b} x}=x\right]$

$\Rightarrow x=e^{-2}$

$\therefore\left\{x: x \in R^{+}\right.$and $\left.f(x)=-2\right\}=\left\{e^{-2}\right\}$

To find: (iii) $f(x y)=f(x)+f(y)$ for all $x, y \in R$

We have,

$f(x y)=\log _{e}(x y)$

$=\log _{e}(x)+\operatorname{loge}(y)$

[Product Rule for Logarithms]

$=f(x)+f(y)\left[\because f(x)=\log _{e} x\right]$

$\therefore f(x y)=f(x)+f(y)$ holds.