1. Log rhythm question help

Consider the following ‘proof’ where for convenience we have numbered each
step.
1. Clearly ( 1)
2
= 1, and so
2. upon taking logarithms we have that ln( 1)
3. Using one of the logarithm rules we have that 2 ln( 1) = 0,
4. or upon division by 2, that ln( 1) = 0.
5. Therefore el
n( 1)
= e0

,
6. which simplifies to 1 = 1.
The conclusion of the ‘proof’ is clearly false, so explain exactly at what
step(s) things went wrong.

So basically I got for the answer: log(-1)(1)=2

I just want to know if my answer is right. I would really appreciate some helps thanks!

2. Re: Log rhythm question help

Originally Posted by OMg
Consider the following ‘proof’ where for convenience we have numbered each
step.
1. Clearly ( 1)
2
= 1, and so
I have no idea what you mean by that.

2. upon taking logarithms we have that ln( 1)
Taking logarithms of what?

3. Using one of the logarithm rules we have that 2 ln( 1) = 0,
I think I am starting to work it out: you had $\displaystyle 1^2= 1$, is that right?

4. or upon division by 2, that ln( 1) = 0.
5. Therefore el
n( 1)
= e0
Once again, I have no idea what this is supposed to mean.

,
6. which simplifies to 1 = 1.
The conclusion of the ‘proof’ is clearly false, so explain exactly at what
step(s) things went wrong.

So basically I got for the answer: log(-1)(1)=2

I just want to know if my answer is right. I would really appreciate some helps thanks!
If you started with $\displaystyle (-1)^2= 1$ then you should know that the logarithm function is only defined for positive x-values. ln(-1) is not defined so nothing you write with it is correct.

3. Re: Log rhythm question help

I beleive the OP is doing this, an old "proof" that 1 = -1:

$\displaystyle (-1)^2 =1$
$\displaystyle \ln (-1)^2 = \ln(1)$
$\displaystyle 2 \ln (-1) =0$
$\displaystyle \ln (-1) =0$
$\displaystyle e^{\ln(-1)}=e^0=1$
$\displaystyle -1 = 1$

The log of a negative number does exist, but it's a complex number: $\displaystyle \ln Z = R + i \theta$, and so $\displaystyle \ln(-1) = i (1+2k)\pi$. The error in the above proof is in ignoring the imaginary part of$\displaystyle \ln(-1)$.