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Math Help - Can somebody teach me about logarithms?

  1. #1
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    Can somebody teach me about logarithms?

    Okay, so I'm just trying to run through this study guide and at the present time I'm on logarithms before it progresses to complex numbers. I just looked at this math latex site I usually go on to try and express the stuff so people know, but I think you'll be able to figure it out. Also, sorry if there any typos in my English...I've got blisters all over my hands and it's even hard to grip a pencil.

    Here are some problems I'd really like explained to me:

    It says to express as a single logarithm.

    1/2logb(x)-logb(y)-logb(z)<br />

    1/3logb(x)-2/3logb(y)

    2log10(3)+4log10(y)-6log10(z)-8log10(t)

    And then these...

    loga\frac{x^2y}{\sqrt[3]{z}<br />
}

    log2 \sqrt{ \frac{a^6b^4}{z}<br />
}

    Suppose to express those in terms of logarithms.

    I mean, I think I might be able to do the first two out of three...I think it's just rearranging...

    01. 1/2logb(x)-logb(y)-logb(z)<br />

    logb(x)^1/2-logb(y)-logb(z)=logb( \sqrt{x}/yz)

    02. 1/3logb(x)-2/3logb(y)

    logx^1/3-logx^1/3=logb[\frac{\sqrt[3]{x}{ \sqrt[3]{y}^2]

    Must somehow have an error in this one.

    logx^1/3-logx^1/3=logb[\frac{\sqrt[3]{x}{ \sqrt[3]{y}^2
    The answer anyway...
    Last edited by A Beautiful Mind; July 10th 2009 at 08:52 AM.
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  2. #2
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    Quote Originally Posted by A Beautiful Mind View Post
    Okay, so I'm just trying to run through this study guide and at the present time I'm on logarithms before it progresses to complex numbers. I just looked at this math latex site I usually go on to try and express the stuff so people know, but I think you'll be able to figure it out. Also, sorry if there any typos in my English...I've got blisters all over my hands and it's even hard to grip a pencil.

    Here are some problems I'd really like explained to me:

    It says to express as a single logarithm.

    1/2logb(x)-logb(y)-logb(z)<br />

    1/3logb(x)-2/3logb(y)

    2log(10)+4log10(y)-6log10(z)-8log10(t)

    And then these...

    loga\frac{x^2y}{\sqrt[3]{z}<br />
}

    log2 \sqrt{ \frac{a^6b^4}{z}<br />
}

    Suppose to express those in terms of logarithms.

    I mean, I think I might be able to do the first two out of three...I think it's just rearranging...

    01. 1/2logb(x)-logb(y)-logb(z)<br />

    logb(x)^1/2-logb(y)-logb(z)=logb( \sqrt{x}/yz)

    02. 1/3logb(x)-2/3logb(y)

    logx^1/3-logx^1/3=logb[\frac{\sqrt[3]{x}{ \sqrt[3]{y}^2]

    Must somehow have an error in this one while trying t

    The answer anyway...
    Hopefully you know the following logarithm rules...

    \log_{a}{a^p} = p

    a^{\log_{a}{p}} = p

    \log{1} = 0

    \log{a} + \log{b} = \log{ab}

    \log{a} - \log{b} = \log{\frac{a}{b}}

    \log{a^p} = p\log{a}


    Using these rules we can simplify your logarithmic expressions...


    1. \frac{1}{2}\log_b{x} - \log_b{y} - \log_b{z}

     = \log_b{x^{\frac{1}{2}}} - (\log_b{y} + \log_b{z})

     = \log_b{x^{\frac{1}{2}}} - \log_b{yz}

     = \log_b{\frac{x^{\frac{1}{2}}}{yz}}.


    2. \frac{1}{3}\log_b{x} - \frac{2}{3}\log_b{y}

     = \log_b{x^{\frac{1}{3}}} - \log_b{y^{\frac{2}{3}}}

     = \log_b{\frac{x^{\frac{1}{3}}}{y^{\frac{2}{3}}}}

     = \log_b{\left(\frac{x}{y^2}\right)^{\frac{1}{3}}}.


    3. I think you've missed some information...


    4. \log_a{\frac{x^2 y}{\sqrt[3]{z}}}

     = \log_a{\frac{x^2 y}{z^{\frac{1}{3}}}}

     = \log_a{x^2 y} - \log_a{z^{\frac{1}{3}}}

     = \log_a{x^2} + \log_a{y} - \frac{1}{3}\log_a{z}

     = 2\log_a{x} + \log_a{y} - \frac{1}{3}\log_a{z}.


    5. \log_2{\sqrt{\frac{a^6 b^4}{z}}}

     = \log_2{\left(\frac{a^6 b^4}{z}\right)^{\frac{1}{2}}}

     = \log_2{\frac{a^3 b^2}{z^{\frac{1}{2}}}}

     = \log_2{a^3 b^2} - \log_2{z^{\frac{1}{2}}}

     = \log_2{a^3} + \log_2{b^2} - \frac{1}{2}\log_2{z}

     = 3\log_2{a} + 2\log_2{b} - \frac{1}{2}\log_2{z}.
    Last edited by Prove It; July 10th 2009 at 09:00 AM.
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    I fixed the third problem.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Hopefully you know the following logarithm rules...

    \log_{a}{a^p} = p

    a^{\log_{a}{p}} = p

    \log{1} = 0

    \log{a} + \log{b} = \log{ab}

    \log{a} - \log{b} = \log{\frac{a}{b}}

    \log{a^p} = p\log{a}


    Using these rules we can simplify your logarithmic expressions...


    1. \frac{1}{2}\log_b{x} - \log_b{y} - \log_b{z}

     = \log_b{x^{\frac{1}{2}}} - (\log_b{y} + \log_b{z})


    e^(i*pi): How did you get to log_b{y} + log_b{z} = log_b{\frac{y}{z}}? Surely it is equal to log_b{yz}



     = \log_b{x^{\frac{1}{2}}} - \log_b{\frac{y}{z}}

     = \log_b{\frac{x^{\frac{1}{2}}y}{z}}.
    For number 1 I get
    log_b{x^{\frac{1}{2}}} - log_b{y} - log_b{z}

    = log_b{x^{\frac{1}{2}}} - (log_b{y} + log_b{z})

    = log_b{x^{\frac{1}{2}}} - log_b{yz}

    = log_b\frac{x^{\frac{1}{2}}}{yz}
    Last edited by e^(i*pi); July 10th 2009 at 08:55 AM. Reason: Added step for extra clarity
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    For number 1 I get
    log_b{x^{\frac{1}{2}}} - log_b{y} - log_b{z}

    = log_b{x^{\frac{1}{2}}} - (log_b{y} + log_b{z})

    = log_b{x^{\frac{1}{2}}} - log_b{yz}

    = log_b\frac{x^{\frac{1}{2}}}{yz}
    Yes you are right...

    Goes to edit post...
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