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Math Help - Elementary Functions Logarithm question

  1. #1
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    Elementary Functions Logarithm question

    I'd like to know why log (x^2)*(y^3); is 2*log x + 3*log y, and not; 6(log x + log y).

    log (x^2)*(y^3) = 2*log x*(y^3)
    2*3*log x*y
    6*(log x + log y)

    I suppose I'm asking about the order in which I am supposed to conduct operations with logarithms. I know this process is wrong but I can not justify my wrongness and the book I am working from does not elaborate much on logarithms.
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  2. #2
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    Re: Elementary Functions Logarithm question

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

    and \log a^b = b\log a

    threrefore

    \displaystyle \log (x^2y^3)  = \log (x^2)+ \log (y^3) = 2\log x+ 3\log y
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  3. #3
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    Re: Elementary Functions Logarithm question

    Quote Originally Posted by bkbowser View Post
    I'd like to know why log (x^2)*(y^3); is 2*log x + 3*log y, and not; 6(log x + log y).

    log (x^2)*(y^3) = 2*log x*(y^3)
    2*3*log x*y
    6*(log x + log y)

    I suppose I'm asking about the order in which I am supposed to conduct operations with logarithms. I know this process is wrong but I can not justify my wrongness and the book I am working from does not elaborate much on logarithms.
    x^2= x*x, y^3= y*y*y so x^2y^3 would be x*x*y*y*y. 6(log x+ log y)= 6log(xy)=log(x^6y^6) and x^6y^6= x*x*x*x*x*x*y*y*y*y*y*y
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    Re: Elementary Functions Logarithm question

    Hello, bkbowser!

    You are taking liberties with Algebra . . .


    \text{I'd lik\,\!e to know why }\,\log (x^2y^3) \:=\:2\log x + 3\log y
    . . \text{and not }6(\log x + \log y).

    \log (x^2y^3) \:=\: 2\log (xy^3) .This is wrong!

    Suppose your answer is correct . . .

    You have: . 6(\log x + \log y) \;=\; 6\log(xy) \;=\;\log(xy)^6 \;=\;\log(x^6y^6)

    . . See?
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