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Math Help - Logarithms with an x,y, and z values?

  1. #1
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    Logarithms with an x,y, and z values?

    Take this one for example
    1/3log7^x +13log7^y - 5log5^z

    How would I reduce this to a single logarithm?
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  2. #2
    MHF Contributor Amer's Avatar
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    you should know that

    log(a)^b=blog(a)

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

    log(a)+log(b)=log(ab)

    apply these two formula you will get the answer
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  3. #3
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    Quote Originally Posted by Amer View Post
    you should know that

    log(a)^b=blog(a)

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

    log(a)+log(b)=log(ab)

    apply these two formula you will get the answer
    If only it was that easy for me. Using that I can get it down to
    log7(x^(1/3)*y^(13)) -5log5^z

    What is throwing me off is the different bases. 5 and 7. And then the different variables x,y, and a third variable z.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by bilbobaggins View Post
    Take this one for example
    1/3log7^x +13log7^y - 5log5^z

    How would I reduce this to a single logarithm?
    see

    1/3 log(7^x)

    log(7^{1/3x})

    13log(7^y)=??
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  5. #5
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    Quote Originally Posted by Amer View Post
    see

    1/3 log(7^x)

    log(7^{1/3x})

    13log(7^y)=??
    Oh, I just realized that before I refreshed and saw your response. Thanks.
    Last edited by bilbobaggins; June 5th 2009 at 02:34 PM.
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  6. #6
    MHF Contributor Amer's Avatar
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    the number the is multiplies with the Logarithm is a power for what is inside it

    see this

    a(log(b))=log(a^b)

    a(log(b)^c)=log(b^{ac})

    so

    \frac{1}{3} log 7^x = log 7^{\frac{x}{3}}

    13 log 7^y = log 7^{13y}

    -5log 5^z=-log 5^{5z}

    now

    log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\left(7^{\frac{x}{3}}\right)\left(7^{13y  }\right)-log 5^{5z}

    log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\dfrac{\left(7^{\frac{x}{3}}\right)\left  (7^{13y}\right)}{5^{5z}}
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  7. #7
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    Quote Originally Posted by Amer View Post
    the number the is multiplies with the Logarithm is a power for what is inside it

    see this

    a(log(b))=log(a^b)

    a(log(b)^c)=log(b^{ac})

    so

    \frac{1}{3} log 7^x = log 7^{\frac{x}{3}}

    13 log 7^y = log 7^{13y}

    -5log 5^z=-log 5^{5z}

    now

    log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\left(7^{\frac{x}{3}}\right)\left(7^{13y  }\right)-log 5^{5z}

    log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\dfrac{\left(7^{\frac{x}{3}}\right)\left  (7^{13y}\right)}{5^{5z}}
    So you can just take the base out of the log and put it in a fraction? Thanks, I didn't know that.
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  8. #8
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    Quote Originally Posted by bilbobaggins View Post
    So you can just take the base out of the log and put it in a fraction?
    No.

    Are you saying that your 7s were actually the bases and the stuff you presented as exponents were really arguments?
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  9. #9
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by bilbobaggins View Post
    Take this one for example
    1/3log7^x +13log7^y - 5log5^z

    How would I reduce this to a single logarithm?
    If those numbers after "log" are the base (ie: \frac{1}{3}log_7(x) + 13log_7(y) - 5log_5(z) then you can combine the first two terms using the laws written by Amer

    log_7(x^{\frac{1}{3}}y^{13})

    Use the change of base rule on this new term and the second term to make them the same base
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  10. #10
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    Quote Originally Posted by e^(i*pi) View Post
    If those numbers after "log" are the base (ie: \frac{1}{3}log_7(x) + 13log_7(y) - 5log_5(z) then you can combine the first two terms using the laws written by Amer

    log_7(x^{\frac{1}{3}}y^{13})

    Use the change of base rule on this new term and the second term to make them the same base
    So (log(x^1/3*y^13)/log7) - (log(z^5)/log5)?
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  11. #11
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    Quote Originally Posted by stapel View Post
    No.

    Are you saying that your 7s were actually the bases and the stuff you presented as exponents were really arguments?
    Yeah pretty much, sorry. I don't know exactly how to put logarithms into a computer.
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  12. #12
    MHF Contributor Amer's Avatar
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    change bases in logarithms

    log_{a}b=\frac{log_c b}{log_c a}

    so you can write

    -5log_{5}z=-log_5 z^5=-\frac{log_{7}(z^5)}{log_{7}(5)}

    to change it base so you can apply what I mention before



    log_{7}(x^{\frac{1}{3}}y^{13})-\frac{log_{7}(z^5)}{\log_{7}(5)}

    {\log_{7}(5)} consider as constant right

    log_{7}(x^{\frac{1}{3}}y^{13})-log_{7}(z^5)(\log_{7}(5))^{-1}


    log_{7}(x^{\frac{1}{3}}y^{13})-log_{7}(z^{5(\log_{7}(5))^{-1}})

    log_{7}\left(\frac{(x^{\frac{1}{3}}y^{13})}{(z^{5(  \log_{7}(5))^{-1}})}\right)
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