# [SOLVED] Extraneous Solutions - Logarithms

• Apr 13th 2008, 07:28 AM
chrozer
[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.
• Apr 13th 2008, 07:38 AM
Mathstud28
Ok
Quote:

Originally Posted by 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.

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
• Apr 13th 2008, 07:41 AM
Moo
Hello mathstud,

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 $\displaystyle \frac{5}{2}<5$
• Apr 13th 2008, 07:44 AM
chrozer
Quote:

Originally Posted by Moo
Hello mathstud,

No..sorry. The question just said what I posted earlier.

MathStud: I have no idea what u did there.
• Apr 13th 2008, 07:49 AM
Moo
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...
• Apr 13th 2008, 07:57 AM
Soroban
Hello, chrozer!

Quote:

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.

• Apr 13th 2008, 07:58 AM
Moo
GRRRRRRRRRRR

I made a sign error ! Pardon me...
• Apr 13th 2008, 08:02 AM
Mathstud28
Ok
Quote:

Originally Posted by chrozer
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 $\displaystyle ln(-1)=\pi{i}$
• Apr 13th 2008, 08:10 AM
chrozer
Quote:

Originally Posted by Mathstud28
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
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?

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

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

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?
• Apr 13th 2008, 08:59 AM
Mathstud28
Ok
Quote:

Originally Posted by chrozer
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?

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

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

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
• Apr 13th 2008, 09:06 AM
chrozer
Quote:

Originally Posted by Mathstud28
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.
• Apr 13th 2008, 09:09 AM
Moo
Quote:

Originally Posted by chrozer
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 ;)
• Apr 13th 2008, 09:41 AM
chrozer
Quote:

Originally Posted by Moo
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.