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Math Help - Logarithmical mess :(

  1. #1
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    Logarithmical mess :(

    So i need some help with two problems this time and the first one is:

    ln910+2ln59ln9

    I have already tried to simplify it according to the logarithmical laws but i just can't get it right, maybe i find this so difficult because it's natural logarithms which i am not used to dealing with.

    The second one is:
    lne3+ln(ee3)+ln1e32


    I keep getting the answer 3 but that is apperantly not correct and i honestly can't see why. I figured that since e=e^1/2I would after simplifiyng ee3 get 3/2. I also figured that ln1e32
    would equal -3/2. That would mean that the whole thing equals 3...but it's not lol. I could really use some help.
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  2. #2
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    Re: Logarithmical mess :(

    Let me correct the numbers, they jumped all over the place

    Number one is

    ln(10/9)+2ln(9/5)-ln9

    Number two:

    ln(e^3)+ln(sqrt e * e^3)+ln (1/ (e^3/2) )
    Last edited by Foxtrot; July 9th 2014 at 09:07 AM.
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  3. #3
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    Re: Logarithmical mess :(

    What do you mean you are getting a mess? Write out your steps.
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    Re: Logarithmical mess :(

    Quote Originally Posted by SlipEternal View Post
    What do you mean you are getting a mess? Write out your steps.
    On the first one i did this:

    ln(10/9)+2ln(9/5)-ln9

    ln(10/ (3^2)+ln(9/5)^2-ln(3^2)

    After this step im blocked.

    Number two i did this:

    ln(e^3)+ln(sqrt e * e^3)+ln (1/ (e^3/2) )

    3+ ln(e^(1/2) * e^3)+3/2

    3+ (3/2) +(3/2) = 3+3=6 (it became 6 this time)
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: Logarithmical mess :(

    Quote Originally Posted by Foxtrot View Post
    ln(10/ (3^2)+ln(9/5)^2-ln(3^2)

    After this step im blocked.
    Remember that lnA + lnB = ln (AxB), and lnA - lnB = ln(A/B). So your problem becomes

    ln((10/9 x (9/5)^2)/9) = ln(10/(5^2)) = ln(10/25) = ln(2/5).

    Quote Originally Posted by Foxtrot View Post
    Number two i did this:

    ln(e^3)+ln(sqrt e * e^3)+ln (1/ (e^3/2) )

    3+ ln(e^(1/2) * e^3)+3/2
    That should be 3+ ln(e^(1/2) * e^3)- 3/2. Note the minus sign!

    Quote Originally Posted by Foxtrot View Post
    3+ (3/2) +(3/2) = 3+3=6 (it became 6 this time)
    Should be 3 + 1/2 + 3 - 3/2 = 5.
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  6. #6
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    Re: Logarithmical mess :(

    Hello, Foxtrot!

    \ln\left(\tfrac{10}{9}\right)+2\ln\left(\tfrac{9}{  5}\right)- \ln(9)

    We have: . \ln\left(\tfrac{10}{9}\right) + \ln\left(\tfrac{9}{5}\right)^2 - \ln(9)

    . . . . . . =\;\ln\left(\tfrac{10}{9}\right) + \ln\left(\tfrac{81}{25}\right) - \ln(9)

    . . . . . . =\;\ln\left(\frac{\frac{10}{9}\cdot\frac{81}{25}}{  9}\right) \;=\;\ln\left(\tfrac{2}{5}\right)




    \ln\left(e^3\right) + \ln\left(\sqrt{e}\cdot e^3\right)+ \ln \left(\frac{1}{e^{\frac{3}{2}}}\right)

    We have: . \ln(e^3) + \ln\left(e^{\frac{1}{2}}\cdot e^3\right) + \ln\left(\frac{1}{e^{\frac{3}{2}}}\right)

    . . . . . . =\;\ln(e^3) + \ln\left(e^{\frac{7}{2}}\right) + \ln\left(\frac{1}{e^{\frac{3}{2}}}\right)

    . . . . . . =\;\ln\left(e^3\cdot e^{\frac{7}{2}}\cdot\frac{1}{e^{\frac{3}{2}}} \right)

    . . . . . . =\;\ln(e^5) \;=\;5
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