Show that the function f : R → R given by f(x) = log a
(x + 
), a > 0, a ≠ 1 is invertible and find its inverse.
Text Solution
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Sol. Let x, y be any two distinct real numbers, then x ≠ y
⇒ x + 
≠ y + 
⇒ log a (x + 
) ≠ log a (y + 
)
⇒ f(x) ≠ f(y)
⇒ f is an injection
Let f(x) = y, then
log a (x + 
) = y
⇒ x + 
= a y ⇒ – x + 
= a
–y ∴ 2x = a
y –a –y ⇒ x = 
Thus for every y ∈ R, there exists x = 
∈ R such that f(x) = y, therefore f is a surjection. Hence f is a bijection or we can say f is one-one onto.
⇒ f is invertible.
x = 
(a
y –a –y )
⇒ f –1 (y) = 
(a y –a –y )
∴ f –1 (x) = 
(a x –a –x )
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