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Thread: solve log equation

  1. #1
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    solve log equation

    Solve the equation and express the answer in terms of $\displaystyle e$.

    $\displaystyle 7+9lnx=9$

    This is what I am guessing:
    $\displaystyle 9lnx=2$
    $\displaystyle lnx={2/-9}$
    $\displaystyle log_e(x)={2/-9}$
    $\displaystyle log^{2/-9}=x$
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  2. #2
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    Quote Originally Posted by theevilp0ptart View Post
    Solve the equation and express the answer in terms of $\displaystyle e$.

    $\displaystyle 7+9lnx=9$

    This is what I am guessing:
    $\displaystyle 9lnx=2$
    $\displaystyle lnx={2/-9}$
    $\displaystyle log_e(x)={2/-9}$
    $\displaystyle log^{2/-9}=x$
    why would it be -9? it was positive just a step ago!
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  3. #3
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    Quote Originally Posted by theevilp0ptart View Post
    Solve the equation and express the answer in terms of $\displaystyle e$.

    $\displaystyle 7+9lnx=9$

    This is what I am guessing:
    $\displaystyle 9lnx=2$
    $\displaystyle lnx={2/{\color{red}-}9}$
    $\displaystyle log_e(x)={2/{\color{red}-}9}$
    $\displaystyle {\color{blue}log^{2/{\color{red}-}9}=x}$
    NO!

    "log" is not a variable. Just like you don't say "sin^2". It has to be the sin OF something. Similarly, you have to take the LOGARITHM of something.

    Recall the definition: $\displaystyle log_{a} b = c \: \: \iff \: \: a^{c} = b$

    The log operator takes the form of $\displaystyle a^{c} = b$ and allows you to solve for the exponent c if you have a and b. Look at the definition and see how it works here.

    And as blair_alane said, where did the negative come from?
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  4. #4
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    So would it be this:

    $\displaystyle 9lnx=2$
    $\displaystyle lnx=2/9$
    $\displaystyle log_{e}x=2/9$
    $\displaystyle e^{2/9}=x$
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  5. #5
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    Hello! Yup, your answer is correct.
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