Question

Let $f:R^{+}\to R$, where $R^{+}$ is the set of all positive real numbers, be such that $f\left(x\right)=lo{g}_{e}x$, Determine
i) The image set of the domain of f
ii) $\{x:f(x)=-2\}$
iii) Whether $f\left(xy\right)=f\left(x\right)+f\left(y\right)$ holds.

Answer 1

We have,
$f=R^{+}\to R$
and $f\left(x\right)=lo{g}_{e}x\dots \dots \left(i\right)$

(i) Now,
$f=R^{+}\to R$
$\therefore$ the image set of the domain of f = R

(ii) Now,
$\{x:f(x)=-2\}$
$\Rightarrow f\left(x\right)=-2\dots \dots \left(ii\right)$
Using equation (i) and equation (ii) we get
$lo{g}_{e}x=-2$
$\Rightarrow x=e^{-2}[\because lo{g}_{a}b=c\Rightarrow b=a^c]$
$\therefore \{x:f(x)=-2\}=\left\{e^{-2}\right\}$

(iii) Now,
$f\left(xy\right)=lo{g}_e\left(xy\right)\left[f\right(x)=lo{g}_{e}x]$
$=lo{g}_{e}x+lo{g}_{e}y[\because lo{g}_{mn}=lo{g}_m+lo{g}_n]$
$f\left(x\right)+f\left(y\right)$
$\therefore f\left(xy\right)=f\left(x\right)+f\left(y\right)$

Yes, $f\left(xy\right)=f\left(x\right)+f\left(y\right)$