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Math Help - Expanding logarithm into sums and difference

  1. #1
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    Expanding logarithm into sums and difference

    Please don't be scared when you see this.

    The problem asks me to expand the following logarithm into sums and differences of logarithms whose arguments are as simple as possible:

    log[((√x2-1)(x2-x)x3)/(x(x-2)104)]

    This is what I did:

    = log(√x2-1)+log(x2-x)+3log(x)-log(x)+log(x-2)+log(104)

    = log(√x2-1)+log(x2-x)+3log(x)-log(x)+log(x-2)+4log(10) as final answer

    Not sure if this is correct or if I followed the steps correctly. Would appreciate if someone would look over this and tell me if I did it right.
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Re: Expanding logarithm into sums and difference

    Quote Originally Posted by FatimaA View Post
    = log(√x2-1)+log(x2-x)+3log(x)-log(x)+log(x-2)+4log(10) as final answer

    Not sure if this is correct or if I followed the steps correctly. Would appreciate if someone would look over this and tell me if I did it right.
    Be very careful with your signs. And we can also go a bit further in the expansion. I'm assuming that that radical is supposed to be \sqrt{x^2-1} and not \left(\sqrt{x^2}-1\right) as you have written:

    \log\left[\frac{\left(\sqrt{x^2-1}\right)\left(x^2-x\right)\left(x^3\right)}{10^4x(x-2)}\right].

    Start by splitting the fraction:

     = \log\left[\left(\sqrt{x^2-1}\right)\left(x^2-x\right)\left(x^3\right)\right]-\log\left[10^4x(x-2)\right]

    Now split the factors:

     = \left[\log\sqrt{x^2-1}+\log\left(x^2-x\right)+\log x^3\right]-\left[\log10^4+\log x+\log(x-2)\right]

    Notice that the entire denominator ends up getting subtracted. That is, we will have to distribute the -1 to each term on the right side:

     = \log\sqrt{x^2-1}+\log\left(x^2-x\right)+\log x^3-\log10^4-\log x-\log(x-2)

    Notice also that x^2-x can be factored as x(x-1), giving us

     = \log\sqrt{x^2-1}+\log x+\log(x-1)+\log x^3-\log10^4-\log x-\log(x-2)

    And finally, remember that \sqrt[n]{a^m}=a^{m/n}. So, we can convert that radical into a rational exponent:

     = \log\left(x^2-1\right)^{1/2}+\log x+\log(x-1)+\log x^3-\log10^4-\log x-\log(x-2)

    Now, can you finish it?
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  3. #3
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    Re: Expanding logarithm into sums and difference

    \log\left[\frac{\sqrt{x^2-1} \cdot (x^2-x) \cdot x^3}{x \cdot (x-2) \cdot 10^4}\right]

    \log\left[\frac{\sqrt{x^2-1} \cdot x(x-1) \cdot x^3}{x \cdot (x-2) \cdot 10^4}\right]

    \log\left[\frac{\sqrt{x-1} \cdot \sqrt{x+1} \cdot (x-1) \cdot x^3}{(x-2) \cdot 10^4}\right]

    \log\left[\frac{(x-1)^{3/2} \cdot (x+1)^{1/2} \cdot x^3}{(x-2) \cdot 10^4}\right]

    \frac{3}{2}\log(x-1) + \frac{1}{2}\log(x+1) + 3\log(x) - \log(x-2) - 4
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  4. #4
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    Re: Expanding logarithm into sums and difference

    Quote Originally Posted by skeeter View Post
    \frac{3}{2}\log(x-1) + \frac{1}{2}\log(x+1) + 3\log(x) - \log(x-2) - 4
    Oh yeah, I forgot to factor the difference of squares there. Thanks skeeter.
    Thanks from FatimaA
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  5. #5
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    Re: Expanding logarithm into sums and difference

    I'm a little confused. You both seemed to do it differently.

    Skeeter how did you get log(x-2)-4?
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  6. #6
    MHF Contributor Reckoner's Avatar
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    Re: Expanding logarithm into sums and difference

    Quote Originally Posted by FatimaA View Post
    I'm a little confused. You both seemed to do it differently.
    Not really. They're the same steps, just in a slightly different order. And I didn't finish the last few steps in my post. But it will come out the same if you continue from there.

    Quote Originally Posted by FatimaA View Post
    Skeeter how did you get log(x-2)-4?
    \log_{10}10^4=4.
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  7. #7
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    Re: Expanding logarithm into sums and difference

    I thought it was supposed to be simplified as 4log(10)?
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  8. #8
    MHF Contributor Reckoner's Avatar
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    Re: Expanding logarithm into sums and difference

    Quote Originally Posted by FatimaA View Post
    I thought it was supposed to be simplified as 4log(10)?
    Yes. And what is \log10?
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  9. #9
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    Re: Expanding logarithm into sums and difference

    Thanks.

    Now I don't know how Skeeter got (3/2)log(x-1). I don't understand things if they are not totally spelled out for me lol. I was confused by the whole 3rd step in Skeeter's solution. How did you get (x-1)^3/2?
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  10. #10
    MHF Contributor Reckoner's Avatar
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    Re: Expanding logarithm into sums and difference

    Quote Originally Posted by FatimaA View Post
    Thanks.

    Now I don't know how Skeeter got (3/2)log(x-1). I don't understand things if they are not totally spelled out for me lol. I was confused by the whole 3rd step in Skeeter's solution. How did you get (x-1)^3/2?
    Well, \sqrt{x-1} = (x-1)^{1/2}. And there's also another factor of x-1. So what is (x-1)^{1/2}\cdot(x-1)? Just add the exponents.
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  11. #11
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    Re: Expanding logarithm into sums and difference

    Ohhhhh okay I see now. Thank you for clearing that up.

    I really love this site.
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