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.
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
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 $\displaystyle \frac{5}{2}<5$
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...
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.
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 $\displaystyle ln(-1)=\pi{i}$.
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?
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