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Math Help - Exponentials and Logs

  1. #1
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    Question Exponentials and Logs

    Hi guys, I have come across this question that I'm not sure how to answer. It goes as follows

    The function f is given by

    f(x) = ln(3x-6) x>2

    a)write down f^(-1)(x) - ie the reverse function

    To do this, our teacher says to make f(x)=y, swap the x and ys, then solve for y

    This gives me
    y = ln(3x-6)
    x = ln(3y-6)
    e^x = 3y-6

    isnt it then simply f^(-1)(x) = (e^x+6)/3

    What do I do now??

    The next question is b) write down the domain and range of f^(-1)(x), which I know seems really simple but im not too sure how to do it

    Isn't the range for a function the domain for the inverse and vice versa?

    Cheers
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  2. #2
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    Re: Exponentials and Logs

    You have done exactly right- that is the formula for f^{-1}(x) and you don't need to do anything else. If you really want to you might 'simplify' that a little as f^{-1}(x)= \frac{1}{3}e^x+ 2.

    Yes, the domain and range swap for the inverse function. Here, the domain of f is x> 2 (actually, that's the "natural domain". If [tex]x\le 2[tex] then 3x- 6\le 0 and ln(x) is only defined for x> 0) and the range is all real number (x very close to 2 will give 3x- 6 very close to 0 and so ln(3x- 6) "very close to -\inty" while if x is arbitrarily large, ln(3x- 6) is "very close to \infty". So the domain of f^{-1}(x) is all real numbers and its range is y> 2.
    Last edited by HallsofIvy; February 17th 2013 at 05:23 AM.
    Thanks from adamcobz
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  3. #3
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    Re: Exponentials and Logs

    Quote Originally Posted by adamcobz View Post
    The function f is given by
    f(x) = ln(3x-6) x>2
    a)write down f^(-1)(x) - ie the reverse function, then simply f^(-1)(x) = (e^x+6)/3

    b) write down the domain and range of f^(-1)(x)
    Good work thus far: f^{-1}(x)=\frac{e^x+6}{3}.

    Clearly the domain is all real numbers and the range must be (2,\infty). Think about it: (\forall x)[e^x>0.
    Thanks from adamcobz
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