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Thread: Find the equation of the inverse function..Logarithm.

  1. #1
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    Find the equation of the inverse function..Logarithm.

    1a)
    $\displaystyle y=log_5 x$
    $\displaystyle x=log_5 y$
    $\displaystyle 5^x=y$
    $\displaystyle y=5^x$
    $\displaystyle f^{-1} (x)=5^x$


    b)
    $\displaystyle y=-log_5 (-x)$
    $\displaystyle x=-log_5 (-y)$
    $\displaystyle -5^x=-y$
    $\displaystyle -y=-5^x$
    $\displaystyle f^{-1} (-x)=-5^x$

    Do those seem correct?
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  2. #2
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    Quote Originally Posted by NotSoBasic View Post
    1a)
    $\displaystyle y=log_5 x$
    $\displaystyle x=log_5 y$
    $\displaystyle 5^x=y$
    $\displaystyle y=5^x$
    $\displaystyle f^{-1} (x)=5^x$


    b)
    $\displaystyle y=-log_5 (-x)$
    $\displaystyle x=-log_5 (-y)$
    $\displaystyle -5^x=-y$
    $\displaystyle -y=-5^x$
    $\displaystyle f^{-1} (-x)=-5^x$

    Do those seem correct?

    I agree with the first question.

    I think the 2nd question should be
    $\displaystyle y=-log_5 (-x)$
    $\displaystyle x=-log_5 (-y)$
    $\displaystyle -x=log_5 (-y)$
    $\displaystyle 5^{-x}=-y$
    $\displaystyle -y=5^{-x}$
    $\displaystyle y=-5^{-x}$
    $\displaystyle f^{-1}=-5^{-x}$
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  3. #3
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    Hello, NotSoBasic!

    The first one is correct . . .



    $\displaystyle b) \;\;y \:=\:-\log_5 (-x)$

    We have: .$\displaystyle x \:=\:-\log_5(-y)$

    Switch sides: .$\displaystyle -\log_5(-y) \:=\:x$

    Multiply by -1: .$\displaystyle \log_5(-y) \:=\:-x$

    Exponentiate: .$\displaystyle -y \:=\:5^{-x}$

    Multiply by -1: .$\displaystyle y \:=\:-5^{-x}$


    Therefore: .$\displaystyle f^{-1}(x) \:=\:-5^{-x}$



    Edit: too slow again . . .
    .
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  4. #4
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    Those make more sense, thanks guys!
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  5. #5
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    Quote Originally Posted by Soroban View Post


    Edit: too slow again . . .

    happens often to me as well.
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