Question -
Answer -
Given f: R+→ R and f(x) = loge x.
(i) the image set of the domain of f
Domain of f = R+ (set of positive real numbers)
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) {x: f(x) = –2}
Given f(x) = –2
loge x = –2
∴ x = e-2 [since, logb a = c ⇒ a = bc]
∴ {x: f(x) = –2} = {e–2}
(iii) Whether f (xy) = f (x) + f (y) holds.
We have f (x) = loge x ⇒ f (y) = loge y
Now, let us consider f (xy)
F (xy) = loge (xy)
f (xy) = loge (x × y) [since, logb (a×c) = logb a + logb c]
f (xy) = loge x + loge y
f (xy) = f (x) + f (y)
∴ the equation f (xy) = f (x) + f (y) holds.