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Math Help - Logarithm

  1. #1
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    Logarithm

    Solve the equation log_{2}{2}-log_{2}{(x+4)}=2-2log_{2}{x}

    The solutions are x=-2, x=4

    We should reject x=-2 otherwise RHS is undefined, am I right?

    What if I change the question to
    log_{2}{2}-log_{2}{(x+4)}=2-log_{2}{x^2}

    Should I reject x=-2?

    Thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by acc100jt View Post
    We should reject x=-2 otherwise RHS is undefined, am I right?
    yes

    What if I change the question to
    log_{2}{2}-log_{2}{(x+4)}=2-log_{2}{x^2}

    Should I reject x=-2?
    no

    but make sure the answer you give works in the ORIGINAL form of the question, by changing the form of something, you sometimes change its domain, which is what happens here
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by acc100jt View Post
    Solve the equation log_{2}{2}-log_{2}{(x+4)}=2-2log_{2}{x}

    The solutions are x=-2, x=4

    We should reject x=-2 otherwise RHS is undefined, am I right?

    What if I change the question to
    log_{2}{2}-log_{2}{(x+4)}=2-log_{2}{x^2}

    Should I reject x=-2?

    Thanks!
    It's a fascinating problem, isn't it?

    The solution (I believe) is that
    log_2(x^2) = 2 \cdot log_2(|x|) \neq 2 \cdot log_2(x)

    Since x = -2 is not forbidden in your x^2 equation, then it is a solution.

    I must admit I've never seen this mentioned in any of my own classes.

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    It's a fascinating problem, isn't it?

    The solution (I believe) is that
    log_2(x^2) = 2 \cdot log_2(|x|) \neq 2 \cdot log_2(x)

    Since x = -2 is not forbidden in your x^2 equation, then it is a solution.

    I must admit I've never seen this mentioned in any of my own classes.

    -Dan
    Indeed! i have never considered something like that before. and i think, maybe, this holds for all even powers. you can use Graph to show that what you have is correct
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    Because \log x^n = n \log x was only proven for x>0. So it is wrong to say \log x^2 = 2\log x because here x<0 and in reality it should be \log x^2 = 2\log |x|.
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    is up to his old tricks again! Jhevon's Avatar
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    however, here, the original question had the form 2 \log_2 x not \log_2 x^2 = 2 \log_2 |x|, so i think the negative answer should be rejected in this case...wait, was that an issue here? or did we agree on that to begin with?
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  7. #7
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    Of course having x=-2 for \log_2x the root must be rejected.

    The original poster changed the question by setting \log_2x^2, so we can accept x=-2.

    TPH said it all.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    however, here, the original question had the form 2 \log_2 x not \log_2 x^2 = 2 \log_2 |x|, so i think the negative answer should be rejected in this case...wait, was that an issue here? or did we agree on that to begin with?
    I think the main issue here is does
    log(x^2) = 2 \cdot log(x)
    for all x? The answer is only for positive x because
    log(x^2) = 2 \cdot log(|x|)

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Of course having x=-2 for \log_2x the root must be rejected.

    The original poster changed the question by setting \log_2x^2, so we can accept x=-2.

    TPH said it all.
    Quote Originally Posted by topsquark View Post
    I think the main issue here is does
    log(x^2) = 2 \cdot log(x)
    for all x? The answer is only for positive x because
    log(x^2) = 2 \cdot log(|x|)

    -Dan
    good. ok, glad to have cleared that up. i guess i never really encountered something like this in simplifying before, since, for instance, if the question did have \ln x^2, i would not have brought the power down in the first place
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    i guess i never really encountered something like this in simplifying before, since, for instance, if the question did have \ln x^2, i would not have brought the power down in the first place
    I commonly bring the 2 down, but I have never encountered a situation (I don't think) where I've needed to be careful about the absolute value. I think the question is a good wake up call.

    -Dan
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