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

1.	Let f: R+→ R, where R+ is the set of all positive real numbers, be such that f(x) = logex. 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, f : R⁺→ R and f(x) = log_ex. 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. Hence, The image set of f is the set of real numbers. Thus, The image set of f = R ii. {x: f(x) = –2} Given, f(x) = –2 ⇒ log_ex = –2 ∴ x = e^-2 [∵ log_ba = c ⇒ a = b^c] Thus, {x: f(x) = –2} = {e –2} iii. whether f(xy) = f(x) + f(y) holds. We have, f(x) = log_ex ⇒ f(y) = log_ey Now, let us consider f(xy). f(xy) = log_e(xy) ⇒ f(xy) = log_e(x × y) [∵ log_b(a×c) = log_ba + log_bc] ⇒ f(xy) = log_ex + log_ey ∴ f(xy) = f(x) + f(y) Hence, 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) = logex. 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,

f : R⁺→ R and f(x) = log_ex.

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.

Hence,

The image set of f is the set of real numbers.

Thus,

The image set of f = R

ii. {x: f(x) = –2}

Given,

f(x) = –2

⇒ log_ex = –2

∴ x = e^-2

[∵ log_ba = c ⇒ a = b^c]

Thus,

{x: f(x) = –2} = {e –2}

iii. whether f(xy) = f(x) + f(y) holds.

We have,

f(x) = log_ex

⇒ f(y) = log_ey

Now,

let us consider f(xy).

f(xy) = log_e(xy)

⇒ f(xy) = log_e(x × y)

[∵ log_b(a×c) = log_ba + log_bc]

⇒ f(xy) = log_ex + log_ey

∴ f(xy) = f(x) + f(y)

Hence,

The equation f(xy) = f(x) + f(y) holds.