Show that the logarithmic function f : R+0→ R given by f(x) = loga x, a > 0 is a bijection.
TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.
So, is One – One function
⇔ a≠b
⇒ f(a)≠f(b) for all
⇔ f(a) = f(b)
⇒ a = b for all
Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.
So, is Surjection iff for each
, there exists
such that f(a) = b
Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.
To Prove : – Logarithmic function f : R + + → R given by f(x) = loga x, a > 0 is a bijection.
Now, f : R0 + → R given by f(x) = loga x, a > 0
Check for Injectivity:
Let x,y be elements belongs to R0 + i.e such that
So, from definition
⇒ f(x) = f(y)
⇒ loga x = loga y
⇒ loga x – loga y = 0
⇒
⇒
⇒ x = y
Hence f is One – One function
Check for Surjectivity:
Let y be element belongs to R i.e. be arbitrary, then
⇒ f(x) = y
⇒ loga x = y
⇒ x = ay
Above value of x belongs to R0 +
Therefore, for all there exist x = ay such that f(x) = y .
Hence, f is Onto function.
Thus, it is Bijective also