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Math Help - Inverse logarithm problem!

  1. #1
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    Question Inverse logarithm problem!

    The directions say "find a formula for f^-1 and verify that (f of f^-1)(x)=(f^1 of f)(x)=x. The problem is f(x)=100/(1+2^-x). Can anybody help me??
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    Re: I need help with an inverse logarithm problem!!

    My preferred method is to set f(x) = y and solve for x in terms of y:

    y = \dfrac{100}{1+2^{-x}}

    Take the reciprocal: \dfrac{1}{y} = \dfrac{1+2^{-x}}{100}

    Multiply both sides by 100: \dfrac{100}{y} = 1+2^{-x}

    Subtract 1 from both sides: 2^{-x} = \dfrac{100}{y} - 1 = \dfrac{100-y}{y}

    It should be relatively easy to solve for x using logarithms. You can pick whichever base you like although I prefer base e (the natural log)
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    Re: I need help with an inverse logarithm problem!!

    But the back of my book gives me the answer f^-1(x)=log base 2(x/100-x). How do I get to that answer?
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    Re: I need help with an inverse logarithm problem!!

    How far can you get from where I left in my previous post?
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    Re: I need help with an inverse logarithm problem!!

    Not very far. I want to get rid of the 2^-x somehow.
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    Re: I need help with an inverse logarithm problem!!

    Quote Originally Posted by ltlbit94 View Post
    Not very far. I want to get rid of the 2^-x somehow.
    Using logarithms to get this is the best method of solving exponential equations like this, perhaps you should revise them when you next get a chance.

    Since the book wants to use base 2 lets go with that: \log_2(2^{-x}) = \log_2 \left(\dfrac{100-y}{y}\right)

    By our log laws we know that \log_a(a) = 1 \text{ and  } \log_c(a^k) = k \log_c(a) so we can rewrite the LHS as \log_2(2^{-x}) = -x \log_2(2) = -x. Do you understand how I used the laws to get to -x ?


    In the equation as whole we have, after dividing by -1: x = -\log_2 \left(\dfrac{100-y}{y}\right)

    Now that we have x in terms of y rewrite x as f^{-1}(x) and y as x:

    f^{-1}(x) = -\log_2 \left(\dfrac{100-x}{x}\right).

    This is the same as your book's answer, just written in a different way. If you want the book's answer use the log power laws on that -1 outside the logarithm and the effect of raising a fraction to power -1
    Last edited by e^(i*pi); August 7th 2011 at 11:20 AM. Reason: tex
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    Re: I need help with an inverse logarithm problem!!

    If you have given:
    y=\frac{100}{1+2^{-x}}
    and you've to find f^{-1}(x) then change x en y:
    x=\frac{100}{1+2^{-y}} \Leftrightarrow 2^{-y}=\frac{100}{x}-1
    Now take the log with base 2 from each side:
    \log_2\left(2^{-y}\right)=\log_2\left(\frac{100-x}{x}\right)
    \Leftrightarrow -y=\log_2\left(\frac{100-x}{x}\right)
    \Leftrightarrow y= -\log_2\left(\frac{100-x}{x}\right)
    \Leftrightarrow y=\log_2\left(\frac{x}{100-x}\right)
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    Re: Inverse logarithm problem!

    Quote Originally Posted by ltlbit94 View Post
    The directions say "find a formula for f^-1 and verify that (f of f^-1)(x)=(f^1 of f)(x)=x. The problem is f(x)=100/(1+2^-x). Can anybody help me??
    From the content of a pm from the OP, it seems this question is part of an assignment that possibly counts towards the student's grade. See rule 6. Thread closed.
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