Simplifying a log expression with identities

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Homework Help Overview

The discussion revolves around simplifying the logarithmic expression ##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|##. Participants are examining the correctness of the simplification process and the implications of undefined values in logarithmic functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the steps taken to simplify the expression and question where potential mistakes may lie. There is a focus on the conditions under which the logarithmic expressions are defined, particularly regarding the values of ##\cos {x}## and ##\sin {x}##.

Discussion Status

There is an ongoing exploration of the conditions that lead to the expression being marked wrong. Some participants suggest that while the simplification appears correct, the domain restrictions need to be clarified. Multiple interpretations of the problem's requirements are being considered.

Contextual Notes

Participants note that the original expression is undefined for certain values of ##x##, specifically where either ##\sin {x}## or ##\cos {x}## equals zero. This raises questions about the validity of the final simplified expression in those contexts.

ProfuselyQuarky
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I was supposed to simplify the expression ##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|## and apparently it’s wrong. Where’s the mistake? Is it not simplified enough or . . . ?

##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|##

##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\frac {\sin {x}}{\cos {x}}\cdot \cos {x}|##

##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\sin {x}|##

##=\ln |\frac {\cos {x}\sin {x}}{\sin {x}}|##

##=\ln |\cos {x}|##

If it’s really obvious, I’ll be happy with a hint. I thought that this was easy, but apparently I was mistaken?
 
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ProfuselyQuarky said:
I was supposed to simplify the expression ##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|## and apparently it’s wrong. Where’s the mistake? Is it not simplified enough or . . . ?

##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|##

##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\frac {\sin {x}}{\cos {x}}\cdot \cos {x}|##

##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\sin {x}|##

##=\ln |\frac {\cos {x}\sin {x}}{\sin {x}}|##

##=\ln |\cos {x}|##

If it’s really obvious, I’ll be happy with a hint. I thought that this was easy, but apparently I was mistaken?

Looks correct though. Note that Ln0 is undefined, so you might want to specify when the equality is correct.
 
Math_QED said:
Looks correct though. Note that Ln0 is undefined, so you might want to specify when the equality is correct.
Well, I was given the problem with no specifications and would it really affect the answer? I mean, this expression doesn’t deal with ##\ln {0}##.
 
So what is the error?
 
ProfuselyQuarky said:
Well, I was given the problem with no specifications and would it really affect the answer? I mean, this expression doesn’t deal with ##\ln {0}##.
You sure about that? You mean cos x is never equal to zero?
 
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SteamKing said:
You sure about that? You mean cos x is never equal to zero?
Oh, I never said cos x never is equal to zero. ##\cos {\frac {\pi}{2}}=0## and ##\cos {\frac {3\pi}{2}}=0## and so on.
 
But would that affect the answer? How did I simplify wrong?
 
ProfuselyQuarky said:
Oh, I never said cos x never is equal to zero. ##\cos {\frac {\pi}{2}}=0## and ##\cos {\frac {3\pi}{2}}=0## and so on.
So what happens to ln |cos x| when x = (2k+1)π/2 ?
 
SteamKing said:
So what happens to ln |cos x| when x = (2k+1)π/2 ?
We get ln 0 which is undefined
 
  • #10
Oh, wait, so I'd have to specify that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.

That's great but there wasn't a single example of any of that in my book which is why I was thinking that my error had something to do with using the identities wrong.
 
  • #11
ProfuselyQuarky said:
Oh, wait, so I'd have to specify that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.

That's great but there wasn't a single example of any of that in my book which is why I was thinking that my error had something to do with using the identities wrong.

The answer is correct, as long as you specify for which values the expression is undefined.
 
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  • #12
ProfuselyQuarky said:
Oh, I never said cos x never is equal to zero. ##\cos {\frac {\pi}{2}}=0## and ##\cos {\frac {3\pi}{2}}=0## and so on.
Frankly. I don't think this is the problem. After all, the original expression isn't defined for values of x which make cos(x)=0, either.

However, there are values of x for which the original expression is undefined, but are defined for ln(cos(x)) . Throw those out.
 
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  • #13
Why do you think your expression is wrong in the first place?
 
  • #14
Math_QED said:
Why do you think your expression is wrong in the first place?
It was marked wrong.
 
  • #15
ProfuselyQuarky said:
Oh, wait, so I'd have to specify that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.

That's great but there wasn't a single example of any of that in my book which is why I was thinking that my error had something to do with using the identities wrong.
I'm not sure I was very clear in my previous post.

I don't think the problem is:
that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.​
Those values of x at which ##\cos {x}=0## are not in the domain of ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## in the first place.

I think that the problem is that ##\ln |\cos {x}|## is defined for some values of x for which ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## is not defined. I think that you must restrict the domain of the answer to eliminate those.
 
  • #16
SammyS said:
I'm not sure I was very clear in my previous post.

I don't think the problem is:
that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.​
Those values of x at which ##\cos {x}=0## are not in the domain of ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## in the first place.

I think that the problem is that ##\ln |\cos {x}|## is defined for some values of x for which ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## is not defined. I think that you must restrict the domain of the answer to eliminate those.

Even then, it shouldn't be marked wrong since the expression itself is correct.
 
  • #17
Math_QED said:
Even then, it shouldn't be marked wrong since the expression itself is correct.
SammyS was right. The initial and final expressions are not identical, as their ranges are different. The final one is not defined when cos(x)=0. The initial one is not defined when either sin(x) or cos(x) is zero.
 

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