# Math Help - [SOLVED] Extraneous Solutions - Logarithms

1. ## [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.

2. ## Ok

Originally Posted by 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.
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

3. 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 :

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

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

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

If $\Delta=0$ the only solution is $\frac{5}{2}$ wich is an extraneous solution as $\frac{5}{2}<5$

4. 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.

5. 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...

6. 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.

7. GRRRRRRRRRRR

I made a sign error ! Pardon me...

8. ## Ok

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 $log_{10}(-3)=\frac{ln(-3)}{ln(10)}=\frac{ln(3)+ln(-1)}{ln(10)}=\frac{ln(3)+\pi{i}}{ln(10)}$ since its well known that $ln(-1)=\pi{i}$

9. 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 $log_{10}(-3)=\frac{ln(-3)}{ln(10)}=\frac{ln(3)+ln(-1)}{ln(10)}=\frac{ln(3)+\pi{i}}{ln(10)}$ since its well known that $ln(-1)=\pi{i}$
Oh ok. I didn't know that $ln(-1)=\pi{i}$.

Originally Posted by Soroban
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?

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

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

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?

10. ## Ok

Originally Posted by chrozer
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?

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

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

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

11. Originally Posted by Mathstud28
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.

12. Originally Posted by chrozer
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

13. 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.