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Math Help - Simplifying log. expressions.

  1. #1
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    Simplifying log. expressions.

    Help. In a hurry.

    ln(15x^3)/y^5

    I dunno how to do the math tags exactly, but the fraction is 15x^3 over y^5
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by mankvill View Post
    Help. In a hurry.

    ln(15x^3)/y^5

    I dunno how to do the math tags exactly, but the fraction is 15x^3 over y^5
    did you mean \ln \left(\frac{15x^3}{y^5}\right)?

    well, you can use this..

    \ln (ab) = \ln a + \ln b

    \ln (a/b) = \ln a - \ln b

    \ln (a^b) = b\ln a

    this must be sufficient.. pls post what you have done..
    Last edited by kalagota; July 7th 2008 at 08:59 PM.
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  3. #3
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    ...I'm so confused.
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  4. #4
    MHF Contributor kalagota's Avatar
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    aww..

    i'll do the first step..
    notice that inside the parenthesis is a ratio.. so you can use the second identity i've given you..

    \ln \left(\frac{5x^3}{y^5}\right) = \ln (5x^3) - \ln (y^5)

    can you do the rest? just give us what you already did..
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  5. #5
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    I honestly have no idea, sorry. I'm drawing a blank. I'm under so much stress right now. x_x
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  6. #6
    MHF Contributor kalagota's Avatar
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    aww.. first, be calm because stress will kill you..
    next, read your notes and recall everything you already know..

    i'll do this example completely.. try to study and try to answer the rest..

    so, we are left with \ln (5x^3) - \ln (y^5)

    \underbrace{\ln 5x^3} - \ln y^5 = \underbrace{\ln 5 + \ln (x^3)} - \ln (y^5) i used the first identity i gave.. for the minuend.

    then \ln 5 + \ln (x^3) - \ln (y^5) = \ln 5 + 3 \ln x - 5 \ln y this time, i used the third identity..

    can you follow it?
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  7. #7
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    i can't follow anything.
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  8. #8
    MHF Contributor kalagota's Avatar
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    i already give the complete solution until to the final answer.. try reading it along with other examples from your books or sources..

    i think, it's late night there. have rested already?
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  9. #9
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    Mankvill: let's have a crash course in logarithms.

    \ln{ab} = \ln{a} + \ln{b}
    If you got multiplication within the ln brackets, you can separate them like that.

    Another example: \ln{abc} = \ln{a} + \ln{b} + \ln{c}

    Another thing, you can also separate the ln if it had division within brackets:
    \ln{\frac{a}{b}} = \ln{a} - \ln{b}

    \ln{\frac{a}{bc}} = \ln{a} - \ln{b} - \ln{c}

    Notice that when they're multiplied, they ln of each value inside is added. When they're divided, it becomes: ln of the numerator minus ln of the denominator.

    Do you follow now? If you do, reread kalagota's answers.
    EDIT: Kalagota, didn't see that you posted this already.
    Last edited by Chop Suey; July 8th 2008 at 01:24 PM.
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  10. #10
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Chop Suey View Post
    ln(a/bc) = ln(a) - ln(bc) = ln(a) - (ln(b) +ln(c)) = ln(a) - ln(b) - ln(c)
    just added something in the middle on how to come on the last expression.
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  11. #11
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Chop Suey View Post
    EDIT: Kalagota, didn't see that you posted this already.
    no problem.
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