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Thread: [SOLVED] Extraneous Solutions - Logarithms

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    [SOLVED] Extraneous Solutions - Logarithms

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

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

    I got up to $\displaystyle x^2 - 5x - 10^a = 0$ , but I don't know what to do next.
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    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 $\displaystyle x$ in this equation:

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

    I got up to $\displaystyle x^2 - 5x - 10^a = 0$ , 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 $\displaystyle x=3$ as an answer that would be extraneous because even though ti would work you woul dhave $\displaystyle log(3-5)=log(-2)=\frac{ln(2)+\pi{i}}{ln(10)}$ and you cant have an imaginary answer
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    Moo
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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 :

    $\displaystyle x^2-5x-10^a=0$

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

    If $\displaystyle \Delta <0$ there are no real solutions.

    If $\displaystyle \Delta=0$ the only solution is $\displaystyle \frac{5}{2}$ wich is an extraneous solution as $\displaystyle \frac{5}{2}<5$
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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 $\displaystyle \Delta >0$, that is to say $\displaystyle 25-4*10^a>0$ -> $\displaystyle 25>4*10^a$ -> $\displaystyle a<\log(25/4)$

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

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

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

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

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

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

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

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

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

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


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

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    Moo
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    GRRRRRRRRRRR

    I made a sign error ! Pardon me...
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    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 $\displaystyle log_{10}(-3)=\frac{ln(-3)}{ln(10)}=\frac{ln(3)+ln(-1)}{ln(10)}=\frac{ln(3)+\pi{i}}{ln(10)}$ since its well known that $\displaystyle ln(-1)=\pi{i}$
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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 $\displaystyle log_{10}(-3)=\frac{ln(-3)}{ln(10)}=\frac{ln(3)+ln(-1)}{ln(10)}=\frac{ln(3)+\pi{i}}{ln(10)}$ since its well known that $\displaystyle ln(-1)=\pi{i}$
    Oh ok. I didn't know that $\displaystyle ln(-1)=\pi{i}$.

    Quote Originally Posted by Soroban View Post
    Hello, chrozer!


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

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

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

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


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

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

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

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

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

    Would this be right?
    Last edited by chrozer; Apr 13th 2008 at 08:41 AM.
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    MHF Contributor Mathstud28's Avatar
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    Ok

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



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

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

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

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

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

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



    Why did you get rid of the $\displaystyle 4 \cdot\ 10^a$ 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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    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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