Question

Let f: R⁺→ R, where R⁺ is the set of all positive real numbers, be such that f(x) = log_e x. Determine
(i) the image set of the domain of f
(ii) {x: f (x) = –2}
(iii) whether f (xy) = f (x) + f (y) holds.

Answer 1

Given as f: R⁺→ R and f(x) = log_e x.

(i) The domain of f = R⁺ (the set of positive real numbers)

As we know the value of logarithm to the base e (natural logarithm) can take all possible real values.

∴ The image set of f = R

(ii) Given as f(x) = –2

log_e x = –2

∴ x = e^-2 [since, log_b a = c ⇒ a = b^c]

∴ {x: f(x) = –2} = {e^–2}

(iii) Here, we have f (x) = log_e x ⇒ f (y) = log_e y

Then, let us consider the f(xy)

F(xy) = log_e (xy)

f(xy) = log_e (x × y) [since, log_b (a×c) = log_b a + log_b c]

f(xy) = log_e x + log_e y

f(xy) = f (x) + f (y)

Thus the equation f(xy) = f(x) + f(y) holds.