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


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