Is ln|sin x| equivalent to ln sin x?

[1MileCrash]

Homework Statement

y = ln | sin x |

Homework Equations

-

The Attempt at a Solution

No attempt at a solution, because it's more of a "which part of this is incorrect" question.

y = ln | sin x |

Would everyone agree that this is equivalent?

$y = ln \sqrt{(sin x^2)}$

Then by properties of logarithms:


$y = \frac{1}{2}ln (sin x^2)$

And again,

$y = \frac{2}{2}ln sin x$

or

$y = ln sin x$

So then, ln|sin x| is equivalent to ln sin x?

Is this true? If so, is it just essentially a visual demonstration of the domain of the logarithm function?

(note that I'm having trouble placing the 2 power on "sin" but I'm sure you all know what I mean, (sin x)^2)

 

[SteveL27]

Homework Statement

y = ln | sin x |

Homework Equations

-

The Attempt at a Solution

No attempt at a solution, because it's more of a "which part of this is incorrect" question.

y = ln | sin x |

Would everyone agree that this is equivalent?

$y = ln \sqrt{(sin x^2)}$

Then by properties of logarithms:


$y = \frac{1}{2}ln (sin x^2)$

And again,

$y = \frac{2}{2}ln sin x$

or

$y = ln sin x$

So then, ln|sin x| is equivalent to ln sin x?

Is this true? If so, is it just essentially a visual demonstration of the domain of the logarithm function?

(note that I'm having trouble placing the 2 power on "sin" but I'm sure you all know what I mean, (sin x)^2)

Go back to the beginning, think about the domain.

 

[1MileCrash]

As far as I can tell, the domain agrees with what I did..

 

[SteveL27]

As far as I can tell, the domain agrees with what I did..

What is the domain?

 

[1MileCrash]

Greater than 0... so ln | x | has a range of all real numbers. So what's wrong with the steps I took above that says ln x = ln | x |?

I'm not sure what you're trying to tell me, a little vague..

 

[SteveL27]

Greater than 0... so ln | x | has a range of all real numbers. So what's wrong with the steps I took above that says ln x = ln | x |?

I'm not sure what you're trying to tell me, a little vague..

Oh. Have you mentioned x > 0 earlier? My mistake if so ... otherwise ... how would we know?

I thought the idea is to show what's wrong with the definition of y ... which is that the original function has a problem with its domain. But if you already know that x > 0 is the domain, then what are you supposed to do with this problem? What question is being asked?

ps ... what about x < 0 ?

 

[I like Serena]

Hi 1MileCrash!

The problem does not put a limit on the domain.
So for instance x = -pi/3 would be an acceptable value.

If you substitute that in the original and also into the final expression, are the results then the same?

Btw, the log function is only defined for positive numbers.

 

[SammyS]

For $y = \frac{1}{2}\ln \sqrt{(sin^2 x)}$, the domain is all reals except fo the zeros of the sine function.

For $y = \frac{2}{2}\,\ln \sqrt{(sin\, x)}$, the domain is all reals for which sin(x) > 0 .

Therefore, these expressions are not equivalent.