Let f: R+→ R, where R+ is the set of all positive real numbers, be

1.	Let f: R+→ R, where R+ is the set of all positive real numbers, be such that f(x) = loge 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» Given as f: R⁺→ R and f(x) = log_ex. (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_ex = –2 ∴ x = e^-2 [since, log_ba = c ⇒ a = b^c] ∴ {x: f(x) = –2} = {e^–2} (iii) Here, we have f (x) = log_ex ⇒ f (y) = log_ey Then, let us consider the f(xy) F(xy) = log_e(xy) f(xy) = log_e(x × y) [since, log_b(a×c) = log_ba + log_bc] f(xy) = log_ex + log_ey f(xy) = f (x) + f (y) Thus the equation f(xy) = f(x) + f(y) holds.

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

Given as f: R⁺→ R and f(x) = log_ex.

(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_ex = –2

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

∴ {x: f(x) = –2} = {e^–2}

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

Then, let us consider the f(xy)

F(xy) = log_e(xy)

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

f(xy) = log_ex + log_ey

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

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