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    Inx^2-In36=Ine
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    (Original post by Cameron45)
    Inx^2-In36=Ine
    What about it?
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    Trying to solve for x but don't know where to start..
    (Original post by RDKGames)
    What about it?
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    (Original post by Cameron45)
    Trying to solve for x but don't know where to start..
    Plenty of ways to go about it.

    You can add \ln 36 to both sides then combine RHS into a single log by using the rule \log a + \log b = \log(ab)
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    remember that ln(e) has a special value
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    The answer is 6. I've tried adding In36 to the other side but I'm not getting an answer of 6
    (Original post by RDKGames)
    Plenty of ways to go about it.



    You can add \ln 36 to both sides then combine RHS into a single log by using the rule \log a + \log b = \log(ab)
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    (Original post by Cameron45)
    The answer is 6. I've tried adding In36 to the other side but I'm not getting an answer of 6
    That's not a correct answer.

    \ln(6^2) - \ln(36) = \ln(36)-\ln(36) = 0 \neq \ln e
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    Oh. It says here that the answer is 6.


    (Original post by RDKGames)
    That's not a correct answer.


    \ln(6^2) - \ln(36) = \ln(36)-\ln(36) = 0 \neq \ln e
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    (Original post by Cameron45)
    Inx^2-In36=Ine
    ln x^2 - ln 36 = ln e
    ln e = 1
    2ln x= 1 ln 36
    x = e^((1 ln 36)/2)
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    (Original post by brainmaster)
    ln e = 1
    ln x^2 = ln 36
    Note quite. You're basically claiming that 1+\ln 36 = \ln 36

    Anyway, no full solutions.
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    (Original post by RDKGames)
    Note quite. You're basically claiming that 1+\ln 36 = \ln 36

    Anyway, no full solutions.
    I deities my post thanks for correcting that I was multiplying instead of adding and I forgot about full solution I'll take care next time
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