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

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Solution

Given:
f : R⁺ → R
and f (x) = log_ex .............(i)

(a) f : R⁺ → R
Thus, the image set of the domain f = R .

(b) {x : f (x) = $-$ 2
⇒ f (x ) = $-$ 2 .....(ii)
From equations (i) and (ii), we get :
log_ex = $-$ 2
⇒ x = $e^{- 2}$
Hence, { x : f (x) = - 2} = { e ^{– 2}} . [Since log_ab = c ⇒ b = a^c]

(c) f (xy) = log_e(xy) {From(i)}
= log_ex + log_ey [Since log_emn = log_e m + log_en]
= f (x) + f (y)
Thus, f (xy) = f (x) + f (y)
Hence, it is clear that f (xy) = f (x) + f (y) holds.

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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

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