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Question
Show that the logarithmic function f : R0+ → R given by f (x) loga x ,a> 0 is a bijection.
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Solution
`f R^+ → R given by f (x) = log_a x , a > 0`
Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
f(x) = f(y)
`log_a x + log _a y`
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R+(domain).
f(x) = y
`log_a x = y`
⇒ `x = a^y in R^+`
So, for every element in the co-domain, there exists some pre-image in the domain.
⇒ f is onto.
Since f is one-one and onto, it is a bijection.
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