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.

Question

Philoid · Accepted 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. Determinei. the image set of the domain of fii. {x: f(x) = –2}iii. whether 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) = log_ex. Determine
i. the image set of the domain of f

ii. {x: f(x) = –2}

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