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how to find out if ln(x-a) is one-to-one function?
Okay, this is what I have so far,
the question is asking to 'Show analytically that the function y = ln(x-a) is one to one
this is how i'm showing the problem
y = b +ln(x-a)
b + ln(x-a) = b + ln(x,-a,)
ln(x-a) = ln(x,-a,)
how do i continue with the ln's and how do they effect the question?
1 Answer
- ?Lv 49 years agoFavorite Answer
The way I would show that the function is one-to-one is that I would show that the inverse is a function.
ln(x-a)=y
log base e (x-a)= y
e^y=x-a
(e^y) + a= x-a+a
e^y + a = x
e^x + a = y
e^x + a= y is a function so y= ln(x-a) is a one to one function