Question

Why do you have to write ln|x| when the domain simply can't refer to negative values?

Answer 1

Also, since we do focus on that, why is it when doing calculations sometimes \[\sqrt(f(x)^2\] is taken to mean f(x) instead of |f(x)} ?

Answer 2

so you mean why we don't write ln(x) instead of ln|x|

well because in the second, the domain is different. x = -1 isn't in the domain of the first.

Answer 3

In the second, sign information was removed by the squaring, so why did we not directly indicate that?

Answer 4

also about your second question, it depends on the function. if x > 0 then sqrt(x^2) = x.

Answer 5

HI!!

Answer 6

\(\Large \ln(|x|)\) is defined on \(( - \infty, \infty)\) while
while \(\Large \ln(x)\) is only defined on the interval \(( 0, \infty)\)
If your x is defined on the former interval then your corresponding function you should use is ln (|x|) and vice versa.

Answer 7

the domain of \(\ln(x)\) is \(x>0\) but the domain of \(\ln(|x|)\) is \(x\neq 0\)

Answer 8

what @DLS said

Answer 9

almost

Answer 10

you still can't take the log of zero, so zero is out of both

Answer 11

Oh, I see! So we're actually creating a new funtion that's a mirror of the right side!

Answer 12

yes

Answer 13

are you involved in calculus now?

Answer 14

for your second question, sometimes it is assumed or arranged that \(f\) is always non-negative, so that \[\sqrt{f^2}=f\]

Answer 15

Yeah.

Answer 16

then you will know that the derivative of \(\ln(x)\) is \(\frac{1}{x}\) and also the derivative of \(\ln(-x)=\frac{1}{x}\) by the chain rule, so frequently people will write \[\int \frac{dx}{x}=\ln(|x|)\]

Answer 17

as an example of \[\sqrt{f^2}=f\] sometimes they write \[\sqrt{\sin^2(x)}=\sin(x)\] but in those cases the domain is usually restricted to \[[0,\frac{\pi}{2}]\] where sine is positive

Answer 18

Just like @misty1212 said, but do pay attention on the function you are dealing with though. For instance,
\[\Large \sqrt{1 - 2\sin^2x} = |\cos x| \neq \cos x \]

Answer 19

yes, it looks weird, not like a sin wave

Answer 20

the downside is mirrored up

Answer 21

but if \[\sqrt(\sin^2(x))\] would never be similar to sin(x), since |sin(x)| is a different function for ever half perod, how is that implicitly dealt with? What's the derivative of |sin x| ? when x is positive, it's simply cos(x), but does the function |sin x| still change according to that?