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Math Help - solve log equation

  1. #1
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    solve log equation

    Solve the equation and express the answer in terms of e.

    7+9lnx=9

    This is what I am guessing:
    9lnx=2
    lnx={2/-9}
    log_e(x)={2/-9}
    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 e.

    7+9lnx=9

    This is what I am guessing:
    9lnx=2
    lnx={2/-9}
    log_e(x)={2/-9}
    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 e.

    7+9lnx=9

    This is what I am guessing:
    9lnx=2
    lnx={2/{\color{red}-}9}
    log_e(x)={2/{\color{red}-}9}
    {\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: log_{a} b = c \: \: \iff \: \: a^{c} = b

    The log operator takes the form of 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:

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