Question

15. Let R^+ be the set of all positive real numbers.
Let f:R^+→ R: f(x) = logex.
Find (i) range (f) (ii) {x : XE R* and f(x) = -2).
(iii) Find out whether f(xy) = f(x) + f(y) for all x, y eR*.​

Answer 1

✪ᴀɴsᴡᴇʀ✪

1)

Here,

$\:\:\:\:\:\:\: f(x) = log_e(x) = ln(x)$

or $\: e^{f(x)} = x ... (1)$

$\to f'(x) = \frac{1}{x}$

$\to f'(x) > 0$

f is a strictly increasing function. So its maxima and minima will occurs at end points of its domain $(0, \infty).$

Now,

$\:\:\:\:\:\: \lim \limits_{x \to 0^+}e^{f(x)} = \lim \limits_{x \to 0^+} x$

$\to \lim \limits_{x \to 0^+}e^{f(x)} = 0^+$

This is possible only if

$\to \lim \limits_{x \to 0^+} f(x) = - \infty$

Also,

$\to \lim \limits_{x \to \infty} f(x) = \infty$

Hence, Range of f(x) is $(-\infty, \infty)$.

2)

Here, $f(x) = - 2$

$\implies ln(x) = - 2$

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

{ $x: x \in \R, f(x) =-2$} = { $e^{-2}$ }

3)

Now,

$\:\:\:\: f(xy) = ln(xy), f(x) = ln(x)$

and $\:\:\:\:\:f(y) = ln(y)$

$\to e^{f(xy)} = xy, e^{f(x)} = x, e^{f(y)} = y$

$\to e^{f(xy)} = e^{f(x)}e^{f(y)}$

$\to e^{f(xy)} = e^{f(x) + f(y)}$

Tanking log on both side, we get

$\to f(xy) = f(x) + f(y)$

Hence Proved