Natural log's and trigonometric identity's. (1 Viewer)

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stf

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1. The problem statement, all variables and given/known data

prove the identity.

ln csc(x) = -ln sin(x)




3. The attempt at a solution

With a calculator I am able to prove it is in fact an identity, but I have NO clue why? In fact I am quite unclear on logarrithms in general as my algebra class just barely covered them and did not mention the natural log's at all except in passing (As "it would be covered in calculus")...So any help is great as to what's going on!
 
31,919
3,891
1. The problem statement, all variables and given/known data

prove the identity.

ln csc(x) = -ln sin(x)




3. The attempt at a solution

With a calculator I am able to prove it is in fact an identity, but I have NO clue why? In fact I am quite unclear on logarrithms in general as my algebra class just barely covered them and did not mention the natural log's at all except in passing (As "it would be covered in calculus")...So any help is great as to what's going on!
ln means logarithm base e, where e is about 2.71828.

What properties of logarithms, in any base, do you know? One of them is the key to this problem, along with one trig identity.
 

stf

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Ok, Revisiting logarithm rules, it looks like it would be ln 1/sin = ln 1 - ln sinx, so I am assuming that ln 1 = 0 which gives the equation...I still don't fully understand why ln 1 equals zero though....I suppose I need to revisit all this old material. Thanks.
 

Mentallic

Homework Helper
3,797
94
Yes, or equivalently, there's another rule that states [tex]ln(a^b)=b*ln(a)[/tex] and in this case [tex]csc(x)=(sin(x))^{-1}[/tex] so b=-1.

Log is the inverse of the exponential e, just like the square root is the inverse of squaring. If we have [tex]y=e^x[/tex] then [tex]x=ln(y)[/tex] such as if we have [tex]y=x^2[/tex] then [tex]x=\sqrt{y}[/tex].

So ln(1)=x, well this is the same as saying [tex]1=e^x[/tex] and for what value of x does [tex]e^x=1[/tex]?
 
Yes, or equivalently, there's another rule that states [tex]ln(a^b)=b*ln(a)[/tex] and in this case [tex]csc(x)=(sin(x))^{-1}[/tex] so b=-1.

Log is the inverse of the exponential e, just like the square root is the inverse of squaring. If we have [tex]y=e^x[/tex] then [tex]x=ln(y)[/tex] such as if we have [tex]y=x^2[/tex] then [tex]x=\sqrt{y}[/tex].

So ln(1)=x, well this is the same as saying [tex]1=e^x[/tex] and for what value of x does [tex]e^x=1[/tex]?
you forgot an important fact

[tex]e^{-ln(x)} = \frac{1}{x}[/tex]

If he/she uses that fact then it explains the rest :)
 

Mentallic

Homework Helper
3,797
94
I did? I showed the OP the two basic rules that he needs and can manipulate those in such a way to get the answer.

[tex]e^{-lnx}=\frac{1}{x}[/tex] isn't so quickly obvious if you're only just starting to learn the rules [tex]e^{lnx}=x[/tex] and [tex]ln(a^b)=b*lna[/tex]

But you can easily get the result by using these rules that they are learning.
 
I did? I showed the OP the two basic rules that he needs and can manipulate those in such a way to get the answer.

[tex]e^{-lnx}=\frac{1}{x}[/tex] isn't so quickly obvious if you're only just starting to learn the rules [tex]e^{lnx}=x[/tex] and [tex]ln(a^b)=b*lna[/tex]

But you can easily get the result by using these rules that they are learning.
Oh well, I am sure that Jedi Hal will find this post and correct us all, but that said

I learned in High School [tex]x^{-1} = \frac{1}{x}[/tex] which implies

[tex]e^{-x} = \frac{1}{e^x}[/tex] thus

[tex]e^{-ln(x)} = \frac{1}{e^{ln(x)}} = \frac{1}{x}[/tex]

Use this property to prove you problem, OP!
 

Mentallic

Homework Helper
3,797
94
Who's Jedi Hal?

Sure, there's lots of ways to do it.

[tex]e^{-ln(x)}=e^{ln(x^{-1})}=x^{-1}=\frac{1}{x}[/tex]

[tex]e^{-ln(x)}=(e^{ln(x)})^{-1}=(x)^{-1}=\frac{1}{x}[/tex]

[tex]e^{-ln(x)}=(e^{-1})^{ln(x)}=\left(\frac{1}{e}\right)^{ln(x)}=\frac{1}{e^{ln(x)}}=\frac{1}{x}[/tex]
 

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