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Math Help - Logarithm Equation

  1. #1
    Lil
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    Post Logarithm Equation

    I have this one. It's looks easy to solve. I'm trying to solve and nothing. I can't get an answer.
    \frac{1}{2}lg(x^2 +x-5)=lg5x+lg\frac{1}{5x}
    ...
    \frac{1}{2}lg(x^2 +x-5)=lg1 Can I divide /:1/2
    Last edited by Lil; October 24th 2010 at 10:41 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Lil View Post
    I have this one. It's looks easy to solve. I'm trying to solve and nothing. I can't get an answer.
    \frac{1}{2}log(x^2 +x-5)=lg5x+lg\frac{1}{5x}
    You need two rules and one fact here:

    First, the fact that \displaystyle \log x is a one-to-one function. So,

    \displaystyle \log A = \log B \implies A = B

    Secondly, the rules:

    \displaystyle n \log X = \log (X^n)

    And \displaystyle \log X + \log Y = \log (XY)


    What can you do with those? I'd apply the rules first before the fact
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  3. #3
    Lil
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    I know this rules... I think, I something do wrong.
    lg(x^2 +x-5)^{\frac{1}{2}}=lg1

    From here I stuck!
    Last edited by Lil; October 24th 2010 at 10:41 AM.
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  4. #4
    Lil
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    Can I do like this?
    lg\sqrt{x^2 +x-5}=lg1
    Last edited by Lil; October 24th 2010 at 10:41 AM.
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  5. #5
    MHF Contributor harish21's Avatar
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    Yes!

    Can you complete the problem now?

    Hint: What is  \mbox{exp(log(x))}??
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  6. #6
    Lil
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    How I solve and try, I can't get D
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  7. #7
    Lil
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    Quote Originally Posted by harish21 View Post
    Yes!

    Hint: What is  \mbox{exp(log(x))}??
    I don't know... So what it is?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Lil View Post
    Can I do like this?
    log\sqrt{x^2 +x-5}=lg1
    Yes! Now look at the fact i gave you

    If log of something = log of something else, then something = something else... hopefully that makes sense when i put it in words


    the other two posters are suggesting you plug each side of the equation into the exponential function and equate them. you get to the same place eventually by doing that. it's easier to just drop the logs if you know that fact.
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  9. #9
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    remember the exponential function is the inverse function of log so think of exp(log x) as ..?
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  10. #10
    MHF Contributor harish21's Avatar
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    If log(a) = log(b), then

    exp(log(a))= exp(log(b)) \implies a=b

    apply this to what you have stated in post no 4!
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  11. #11
    Lil
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    Quote Originally Posted by Jhevon View Post
    Yes! Now look at the fact i gave you

    If log of something = log of something else, then something = something else... hopefully that makes sense when i put it in words


    the other two posters are suggesting you plug each side of the equation into the exponential function and equate them. you get to the same place eventually by doing that. it's easier to just drop the logs if you know that fact.
    Sorry guys, I correct my mistake. Here aren't log, It's just lg.
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Lil View Post
    Sorry guys, I correct my mistake. Here aren't log, It's just lg.
    The same rules apply

    In fact, if you look at the posts of the others, they were assuming it was ln, i wasn't. The fact works in general.

    \displaystyle \log_a X = \log_a Y \implies a^{ \log_a X} = a^{\log_a Y} \implies X = Y

    Here, \displaystyle a > 1 is any base. For ln(x), \displaystyle a = e, but as you see, you get the same result. Just drop the logs. You can do this because the logarithm function is 1-1
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  13. #13
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    Quote Originally Posted by Lil View Post
    Sorry guys, I correct my mistake. Here aren't log, It's just lg.
    It doesn't matter, the rule in posts 2 and 10 holds for all bases except 0 and 1
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  14. #14
    Lil
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    Quote Originally Posted by Jhevon View Post
    Just drop the logs.
    x^2 +x-5=1
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  15. #15
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Lil View Post
    x^2 +x-5=1
    Yes, now solve that quadratic.

    (Always check if your answers are in the domains of the functions with problems like these. And please remember there is a specific reason why you can just "drop" the logs here. It is not true that you can do that with functions in general, only 1 - 1, that is, invertible ones)
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