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Math Help - [SOLVED] Extraneous Solutions - Logarithms

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    [SOLVED] Extraneous Solutions - Logarithms

    What kinds of solutions would be extraneous when solving for x in this equation:

    \log x + \log (x-5) = a.

    I got up to x^2 - 5x - 10^a = 0 , but I don't know what to do next.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by chrozer View Post
    What kinds of solutions would be extraneous when solving for x in this equation:

    \log x + \log (x-5) = a.

    I got up to x^2 - 5x - 10^a = 0 , but I don't know what to do next.
    I am not saying this is a solution because it is not but say you got x=3 as an answer that would be extraneous because even though ti would work you woul dhave log(3-5)=log(-2)=\frac{ln(2)+\pi{i}}{ln(10)} and you cant have an imaginary answer
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    Hello mathstud,

    And what about a ?

    chrozer : do you have more info about a ?

    To get the solutions, the real solutions i mean (we'll assume this all along), it has to be such as x-5>0 that is to say x>5.

    Study the roots of the trinomial :

    x^2-5x-10^a=0

    \Delta=\frac{25-4*10^a}{2}

    If \Delta <0 there are no real solutions.

    If \Delta=0 the only solution is \frac{5}{2} wich is an extraneous solution as \frac{5}{2}<5
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    Quote Originally Posted by Moo View Post
    Hello mathstud,

    And what about a ?

    chrozer : do you have more info about a ?
    No..sorry. The question just said what I posted earlier.

    MathStud: I have no idea what u did there.
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    Moo
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    I edited my post

    And i continue it.

    If \Delta >0, that is to say 25-4*10^a>0 -> 25>4*10^a -> a<\log(25/4)

    The solutions will be x_1=\frac{5-\sqrt{25-4*10^a}}{2} and x_2=\frac{5+\sqrt{25-4*10^a}}{2}

    x_1 is clearly inferior to 5 as \sqrt{25-4*10^a}>0 -> extraneous solution

    For x_2, we can discuss about it...
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    Hello, chrozer!

    What kinds of solutions would be extraneous when solving for x in this equation:

    \log x + \log (x-5) = a.

    I got up to x^2 - 5x - 10^a = 0 , but I don't know what to do next.

    Quadratic Formula: . x \;=\;\frac{5 \pm\sqrt{25 + 4\!\cdot\!10^a}}{2}

    There are two roots, one of them is: . x_2\:=\:\frac{5 \,{\color{red}-}\, \sqrt{25 + 4\!\cdot\!10^a}}{2}

    The radical is: . r \:=\:\sqrt{25 + 4\!\cdot\!10^a} \:>\:\sqrt{25} \:=\:5 \quad\Rightarrow\quad r \:>\:5

    This makes x_2 negative, an extraneous root.
    . . [We cannot have the log of a negative quantity.]


    Hence, the only root is: . x_1\;=\;\frac{5 \,{\color{red}+}\, \sqrt{25 + 4\!\cdot\!10^a}}{2}
    . . The other is always extraneous.

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  7. #7
    Moo
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    GRRRRRRRRRRR

    I made a sign error ! Pardon me...
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by chrozer View Post
    No..sorry. The question just said what I posted earlier.

    MathStud: I have no idea what u did there.
    What I was doing was not showing you the extraneous answer in this problem but what would make an extraneous answer...because if we got a number x such that x-5<0 then we would have a problem because we would have lets say x=-3 then we would have log_{10}(-3)=\frac{ln(-3)}{ln(10)}=\frac{ln(3)+ln(-1)}{ln(10)}=\frac{ln(3)+\pi{i}}{ln(10)} since its well known that ln(-1)=\pi{i}
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    Quote Originally Posted by Mathstud28 View Post
    What I was doing was not showing you the extraneous answer in this problem but what would make an extraneous answer...because if we got a number x such that x-5<0 then we would have a problem because we would have lets say x=-3 then we would have log_{10}(-3)=\frac{ln(-3)}{ln(10)}=\frac{ln(3)+ln(-1)}{ln(10)}=\frac{ln(3)+\pi{i}}{ln(10)} since its well known that ln(-1)=\pi{i}
    Oh ok. I didn't know that ln(-1)=\pi{i}.

    Quote Originally Posted by Soroban View Post
    Hello, chrozer!


    Quadratic Formula: . x \;=\;\frac{5 \pm\sqrt{25 + 4\!\cdot\!10^a}}{2}

    There are two roots, one of them is: . x_2\:=\:\frac{5 \,{\color{red}-}\, \sqrt{25 + 4\!\cdot\!10^a}}{2}

    The radical is: . r \:=\:\sqrt{25 + 4\!\cdot\!10^a} \:>\:\sqrt{25} \:=\:5 \quad\Rightarrow\quad r \:>\:5

    This makes x_2 negative, an extraneous root.
    . . [We cannot have the log of a negative quantity.]


    Hence, the only root is: . x_1\;=\;\frac{5 \,{\color{red}+}\, \sqrt{25 + 4\!\cdot\!10^a}}{2}
    . . The other is always extraneous.
    Why did you get rid of the 4 \cdot\ 10^a in the equation in the radical part?

    I also asked someone else about this question and he explained it this way:

    \log x + \log (x-5) =\ log [ x(x-5) ] =a
    x(x-5) = 10^a
    x^2-5x-10^a=0

    ax^2+bx+c=0
    a=1, b=-5 , c=10^a

    If the roots of this equation were to be real,
    b^2-4ac \geq 0
    25-4(10^a) \geq 0
    4(10^a) \leq 25
    10^a \leq6.25
    a \leq log (6.25)
    If a > log(6.25), the solution wouldn't be real.

    Would this be right?
    Last edited by chrozer; April 13th 2008 at 09:41 AM.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by chrozer View Post
    Oh ok. I didn't know that ln(-1)=\pi{i}.



    Why did you get rid of the 4 \cdot\ 10^a in the equation in the radical part?

    I also asked someone else about this question and he explained it this way:

    \log x + \log (x-5) =\ log [ x(x-5) ] =a
    x(x-5) = 10^a
    x^2-5x-10^a=0

    ax^2+bx+c=0
    a=1, b=-5 , c=10^a

    If the roots of this equation were to be real,
    b^2-4ac \geq 0
    25-4(10^a) \geq 0
    4(10^a) \leq 25
    10^a \leq6.25
    a \leq log (6.25)
    If a > log(6.25), the solution wouldn't be real.

    Would this be right?
    NO...because this only accounts to the extraneous solutions caused by making \sqrt{u(x)} imaginary it does not account for making \ln(u(x)) undefined
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  11. #11
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    Quote Originally Posted by Mathstud28 View Post
    NO...because this only accounts to the extraneous solutions caused by making \sqrt{u(x)} imaginary it does not account for making \ln(u(x)) undefined
    Ok thnx alot.
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  12. #12
    Moo
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    Quote Originally Posted by chrozer View Post
    Oh ok. I didn't know that ln(-1)=\pi{i}.



    Why did you get rid of the 4 \cdot\ 10^a in the equation in the radical part?
    Because he wants to compare it to 5, as the equation is defined if and only if x>5.

    As 4*10^a is always positive, 25+4*10^a is always superior to 25, so sqrt(25+4*10^a) is always superior to sqrt(25)=5.

    Hence x1 is always superior to (5+5)/2=5
    And x2 is always inferior to (5-5)/2=0

    So the first one is a real solution, the second one an extraneous solution
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  13. #13
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    Quote Originally Posted by Moo View Post
    Because he wants to compare it to 5, as the equation is defined if and only if x>5.

    As 4*10^a is always positive, 25+4*10^a is always superior to 25, so sqrt(25+4*10^a) is always superior to sqrt(25)=5.

    Hence x1 is always superior to (5+5)/2=5
    And x2 is always inferior to (5-5)/2=0

    So the first one is a real solution, the second one an extraneous solution
    Oh...I see. Thnx alot.
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