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Homework Help: Inverse of Logarithm

  1. Nov 4, 2008 #1
    Run into trouble...again.
    This concerns the inverse function of a logarithm
    If a function maps x on to logax, then the inverse maps logax on to x.
    So, f(x) = logax, can be presented as y=logax; therefore, x=ay.
    The book states that the inverse is ax, why is this the inverse?
    I tried determining the inverse by using a basic inverse:
    If x=2
    Then, f(x)=3*2+1=7
    and f-1(x)=(7-1)/3=2
    If the product of logax=y, going the other way involves ay=x, yet the inverse is, apparently, ax.
    I don't see why this is so...

    P.S. Happy voting!
  2. jcsd
  3. Nov 4, 2008 #2


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    Staff: Mentor

    You are correct. I think the book is just redefining "x". Kind of the way you did when you used x in both the function and inverse function definitions....

    It's better to be explicit about a different "x" and "y" for the two different directions...
  4. Nov 4, 2008 #3


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    Homework Helper

    Your work is correct; the difference is mostly a matter of notation.

    One old1 trick for finding inverses of functions is to begin by reversing the roles of [tex] x [/tex] and [tex] y [/tex] at the start. The complete process would look like this.

    f(x) & = \log_a x \\
    y & = \log_a x \tag{Now switch variables}\\
    x & =\log_a y\\
    a^x & = y,

    so the inverse function has equation

    y = a^x

    We are accustomed to writing functions in the [tex] y = [/tex] form in early math courses.

    Footnote 1: I know this is an old procedure because I learned it when I was in high school: if it has been around that long, it is indeed old.
  5. Nov 4, 2008 #4
    Thanks for the quick replies! The responses are much appreciated, and even tickled me :rofl:, not something one would associate with a maths help forum!!!!
    So ax is really ay 'in disguise', since it is the inverse function the y is now the x?
  6. Nov 5, 2008 #5


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    Science Advisor

    Don't write "x" and "y" in isolation. That is what is confusing you. The functions f(x)= ln(x) or f(y)= ln(y) or f(z)= ln(z) are all exactly the same function.

    If f(x)= ln(x) then f-1(x)= ex or f-1(y)= ey or f-1(a)= ea, etc. it doesn't matter what you call the variable.
  7. Nov 5, 2008 #6
    Okay, I thought I had it...
    If f(x)=y, then f(x)=logax, and ay=x, and so af-1(x)=x?
    If f(x)=logax and the inverse is ax, and I attempt an equation it doesn't work.... x has to change going in the different 'ends', and so x in the new equation (the inverse) is y on the first equation...
    I appreciate this may be getting tiresome to explain... but please bear with me! :smile:
  8. Nov 5, 2008 #7


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    Science Advisor

    Sure it does
    [tex]a^{log_a x}= x[/tex]
    [tex]log_a(a^x)= x[/tex]

    Once again, f(x)= loga(x), f(y)= loga(y), f(z)= loga(z), etc. are all exactly the same function.
  9. Nov 5, 2008 #8


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    Homework Helper

    The notation is getting to you.

    f(x) & = \log_a x\\
    y & = \log_a x \\
    x & = \log_a y\\
    a^x & = y

    This means that the function [tex] f(x) = \log_a x [/tex] has as its inverse

    f^{-1}(x) = a^x
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