Let R+ be the set of all positive real numbers. Let f : R+→ R : f(x)

1.	Let R+ be the set of all positive real numbers. Let f : R+→ R : f(x) = logex. Find (i) Range (f) (ii) {x : x ϵ R+ and f(x) = -2}. (iii) Find out whether f(x + y) = f(x). f(y) for all x, y ϵ R.
Answer» Given that f: R+→ R such that f(x) = log_ex To find: (i) Range of f Here, f(x) = log_ex We know that the range of a function is the set of images of elements in the domain. ∴ The image set of the domain of f = R Hence, the range of f is the set of all real numbers. To find: (ii) {x : x ϵ R⁺ and f(x) = -2} We have, f(x) = -2 …(a) And f(x) = log_ex …(b) From eq. (a) and (b), we get log_ex = -2 Taking exponential both the sides, we get ⇒ e^loge^x = e ^-2 [ ∵ Inverse property I . e b^logbx = x] ⇒ x = e-2 ∴{x : x ϵ R⁺ and f(x) = -2} = {e^-2 } To find: (iii) f(xy) = f(x) + f(y) for all x, y ϵ R We have, f(xy) = loge(xy) = loge(x) + loge(y) [Product Rule for Logarithms] = f(x) + f(y) [∵f(x) = logex] ∴ f(xy) = f(x) + f(y) holds.

Let R+ be the set of all positive real numbers. Let f : R+→ R : f(x) = logex. Find (i) Range (f) (ii) {x : x ϵ R+ and f(x) = -2}. (iii) Find out whether f(x + y) = f(x). f(y) for all x, y ϵ R.

Answer»

Given that f: R+→ R such that f(x) = log_ex

To find: (i) Range of f

Here, f(x) = log_ex

We know that the range of a function is the set of images of elements in the domain.

∴ The image set of the domain of f = R

Hence, the range of f is the set of all real numbers.

To find: (ii) {x : x ϵ R⁺ and f(x) = -2}

We have, f(x) = -2 …(a)

And f(x) = log_ex …(b)

From eq. (a) and (b), we get

log_ex = -2

Taking exponential both the sides, we get

⇒ e^loge^x = e ^-2

[ ∵ Inverse property I . e b^logbx = x]

⇒ x = e-2 ∴{x : x ϵ R⁺ and f(x) = -2} = {e^-2 }

To find: (iii) f(xy) = f(x) + f(y) for all x, y ϵ R

We have,

f(xy) = loge(xy)

= loge(x) + loge(y)

[Product Rule for Logarithms]

= f(x) + f(y) [∵f(x) = logex]

∴ f(xy) = f(x) + f(y) holds.