Simplifying ln: A Calculus Struggle

[Tweedybird]

Homework Statement

Simplify
ln [(e^1-3ln2)/(2∏/e)]

Homework Equations

I was using the equation e^lnx = x to try and simplify the numerator, but I am unsure if that is correct.

The Attempt at a Solution

I am very rusty with my calculus and when it comes to using ln, I tend to get stumped with the question.
I had simplified the numerator to -5 using e^lnx = x, but I was very unsure if this is correct; I may be making up rules of my own.
The denominator was simplified to 2∏(e^-1)
The entire fraction was then ln (-5/(2∏(e^-1)) and I couldn't figure out what to do next with this.

 

[SammyS]

Homework Statement

Simplify
ln [(e^1-3ln2)/(2∏/e)]

Homework Equations

I was using the equation e^lnx = x to try and simplify the numerator, but I am unsure if that is correct.

The Attempt at a Solution

I am very rusty with my calculus and when it comes to using ln, I tend to get stumped with the question.
I had simplified the numerator to -5 using e^lnx = x, but I was very unsure if this is correct; I may be making up rules of my own.
The denominator was simplified to 2∏(e^-1)
The entire fraction was then ln (-5/(2∏(e^-1)) and I couldn't figure out what to do next with this.

Hello Tweedybird. Welcome to PF !

If what you have is indeed

[itex]\displaystyle \ln \left(\frac{e^1-3\ln(2)}{2\pi/e}\right)[/itex]​

then you are very far off.

Perhaps you meant

[itex]\displaystyle \ln \left(\frac{e^{1-3\ln(2)}}{2\pi/e}\right)\ .[/itex]​

Well ... there's still a problem.

 

[Tweedybird]

Hello Tweedybird. Welcome to PF !

If what you have is indeed

[itex]\displaystyle \ln \left(\frac{e^1-3\ln(2)}{2\pi/e}\right)[/itex]​

then you are very far off.

Perhaps you meant

[itex]\displaystyle \ln \left(\frac{e^{1-3\ln(2)}}{2\pi/e}\right)\ .[/itex]​

Well ... there's still a problem.

The second one is exactly the question I need help with! Sorry for the confusion, but I'm not sure why there is a problem with it, it is in my textbook as a question to be simplified!

 

[SammyS]

The second one is exactly the question I need help with! Sorry for the confusion, but I'm not sure why there is a problem with it, it is in my textbook as a question to be simplified!

Review properties of logarithms & exponents.

One property that you may find helpful is:

[itex]\displaystyle \ln\left(\frac{A}{B}\right)=\ln(A)-\ln(B)[/itex]​

 

[Tweedybird]

Review properties of logarithms & exponents.

One property that you may find helpful is:

[itex]\displaystyle \ln\left(\frac{A}{B}\right)=\ln(A)-\ln(B)[/itex]​

I figured the question out! A little review and that hint helped me out quite a bit! Like I said in my post, I am a little rusty with this subject! Thank you for your help :)