1. Logarithm

Solve the equation $\displaystyle log_{2}{2}-log_{2}{(x+4)}=2-2log_{2}{x}$

The solutions are $\displaystyle x=-2, x=4$

We should reject $\displaystyle x=-2$ otherwise RHS is undefined, am I right?

What if I change the question to
$\displaystyle log_{2}{2}-log_{2}{(x+4)}=2-log_{2}{x^2}$

Should I reject $\displaystyle x=-2$?

Thanks!

2. Originally Posted by acc100jt
We should reject $\displaystyle x=-2$ otherwise RHS is undefined, am I right?
yes

What if I change the question to
$\displaystyle log_{2}{2}-log_{2}{(x+4)}=2-log_{2}{x^2}$

Should I reject $\displaystyle x=-2$?
no

but make sure the answer you give works in the ORIGINAL form of the question, by changing the form of something, you sometimes change its domain, which is what happens here

3. Originally Posted by acc100jt
Solve the equation $\displaystyle log_{2}{2}-log_{2}{(x+4)}=2-2log_{2}{x}$

The solutions are $\displaystyle x=-2, x=4$

We should reject $\displaystyle x=-2$ otherwise RHS is undefined, am I right?

What if I change the question to
$\displaystyle log_{2}{2}-log_{2}{(x+4)}=2-log_{2}{x^2}$

Should I reject $\displaystyle x=-2$?

Thanks!
It's a fascinating problem, isn't it?

The solution (I believe) is that
$\displaystyle log_2(x^2) = 2 \cdot log_2(|x|) \neq 2 \cdot log_2(x)$

Since x = -2 is not forbidden in your $\displaystyle x^2$ equation, then it is a solution.

I must admit I've never seen this mentioned in any of my own classes.

-Dan

4. Originally Posted by topsquark
It's a fascinating problem, isn't it?

The solution (I believe) is that
$\displaystyle log_2(x^2) = 2 \cdot log_2(|x|) \neq 2 \cdot log_2(x)$

Since x = -2 is not forbidden in your $\displaystyle x^2$ equation, then it is a solution.

I must admit I've never seen this mentioned in any of my own classes.

-Dan
Indeed! i have never considered something like that before. and i think, maybe, this holds for all even powers. you can use Graph to show that what you have is correct

5. Because $\displaystyle \log x^n = n \log x$ was only proven for $\displaystyle x>0$. So it is wrong to say $\displaystyle \log x^2 = 2\log x$ because here $\displaystyle x<0$ and in reality it should be $\displaystyle \log x^2 = 2\log |x|$.

6. however, here, the original question had the form $\displaystyle 2 \log_2 x$ not $\displaystyle \log_2 x^2 = 2 \log_2 |x|$, so i think the negative answer should be rejected in this case...wait, was that an issue here? or did we agree on that to begin with?

7. Of course having $\displaystyle x=-2$ for $\displaystyle \log_2x$ the root must be rejected.

The original poster changed the question by setting $\displaystyle \log_2x^2,$ so we can accept $\displaystyle x=-2.$

TPH said it all.

8. Originally Posted by Jhevon
however, here, the original question had the form $\displaystyle 2 \log_2 x$ not $\displaystyle \log_2 x^2 = 2 \log_2 |x|$, so i think the negative answer should be rejected in this case...wait, was that an issue here? or did we agree on that to begin with?
I think the main issue here is does
$\displaystyle log(x^2) = 2 \cdot log(x)$
for all x? The answer is only for positive x because
$\displaystyle log(x^2) = 2 \cdot log(|x|)$

-Dan

9. Originally Posted by Krizalid
Of course having $\displaystyle x=-2$ for $\displaystyle \log_2x$ the root must be rejected.

The original poster changed the question by setting $\displaystyle \log_2x^2,$ so we can accept $\displaystyle x=-2.$

TPH said it all.
Originally Posted by topsquark
I think the main issue here is does
$\displaystyle log(x^2) = 2 \cdot log(x)$
for all x? The answer is only for positive x because
$\displaystyle log(x^2) = 2 \cdot log(|x|)$

-Dan
good. ok, glad to have cleared that up. i guess i never really encountered something like this in simplifying before, since, for instance, if the question did have $\displaystyle \ln x^2$, i would not have brought the power down in the first place

10. Originally Posted by Jhevon
i guess i never really encountered something like this in simplifying before, since, for instance, if the question did have $\displaystyle \ln x^2$, i would not have brought the power down in the first place
I commonly bring the 2 down, but I have never encountered a situation (I don't think) where I've needed to be careful about the absolute value. I think the question is a good wake up call.

-Dan