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? asked in Education & ReferenceHomework Help · 9 years ago

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

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  • ?
    Lv 4
    9 years ago
    Favorite 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

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