Let f :


Let f : R+ → R, where R+ is the set of all positive real numbers, such that f(x) = loge x. Determine

(a) the image set of the domain of f

(b) {x : f(x) = −2

(c) whether f(xy) = f(x) : f(y) holds



f : R+ → R

and (x) = logex .............(i)

(a) f : R+ → R

Thus, the image set of the domain f = .

(b) {x : f (x) = -">-2

⇒ (x ) = -">2    .....(ii)

From equations (i) and (ii), we get :

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

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

Hence, $\{x: f(x)=-2\}=\left\{e^{-2}\right\} . \quad\left[\right.$ Since $\left.\log _{a} b=c \Rightarrow b=a^{c}\right]$

(c) (xy) = loge(xy)      {From(i)}

 = logex + logey         [Since logemn = loge m + logen]  

 = f (x) + f (y)

Thus, f (xy) = f (x) + f (y)

Hence, it is clear that f (xy) = (x) + f (y) holds.

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