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Thread: exponential and logarithmic functions

  1. #1
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    exponential and logarithmic functions

    I am working on converting logarithmic equations in exponential form but I do not understand how my book explains it.

    The two problems I am having trouble with are:

    ln e^3 = 3

    ln 1/e^2 = -2

    Could someone please explain how to do this so I understand the concept?
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  2. #2
    MHF Contributor Matt Westwood's Avatar
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    Suppose you have an exponential equation like:
    $\displaystyle a^b = c$

    The "logarithm" function says: "What number do I have to raise $\displaystyle a$ to so as to get $\displaystyle c$?"

    So we have $\displaystyle \log_a c = b$.

    Now when the number being raised to the power is $\displaystyle e$ (which is of course 2.71828 ...), the same thing applies.

    So $\displaystyle e^x = y$ is another way of saying $\displaystyle \log_e y = x$. And a special symbol for $\displaystyle \log_e$ is $\displaystyle \ln$.

    Let's look at $\displaystyle \ln e^3 = x$, where we're trying to find $\displaystyle x$.

    This is saying: "What number do I raise $\displaystyle e$ to so as to get $\displaystyle e^3$?"

    What number do I raise $\displaystyle n$ to (where $\displaystyle n$ is any number) to get $\displaystyle n^3$?

    As you can see, $\displaystyle \ln{}$ "undoes" the work that "$\displaystyle e$ to the power of" does.

    That is, logarithm and exponential (if it's the same base) are inverse functions.

    Now the second one is trickier.

    $\displaystyle \ln \left({\frac 1 {e^2}}\right) = x$

    The thing here is to remember that $\displaystyle \ln \left({\frac 1 {x}}\right) = - \ln x$ (it just is, the book probably shows why).

    So $\displaystyle \ln \left({\frac 1 {e^2}}\right) = -\ln e^2 = -2$ (from what we did before).
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  3. #3
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    Thanks

    Wow!

    Thank you very much, that explanation was so much better than my book.

    I appreciate it.
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