How does this simplify to give this answer?

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  • #1
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1. I am on the very last step of this problem, but I don't see how this equation simplifies from
2ln|(2/√(2)) + 1| - 2ln|1 + 0| to 2ln|√(2) + 1|.


2. I thought that since ln|1| = 0, then 2x0=0. And then the answer would be:
2ln|2/√(2) +1|.

I don't see how the 2 in the (2/√(2)) was canceled out to give just √(2)...

Would you please explain this to me?
Thank you very much! :D
 

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  • #2
Dick
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1. I am on the very last step of this problem, but i don't see how this equation simplifies from
2ln|(2/√(2)) + 1| - 2ln|1 + 0| to 2ln|√(2) + 1|.


2. I thought that since ln|1| = 0, then 2x0=0. And then the answer would be:
2ln|2/√(2) +1|.

I don't see how the 2 in the (2/√(2)) was canceled out to give just √(2)...

Would you please explain this to me?
Thank you very much! :d
2=(√2)*(√2). Or 2=2^1, √2=2^(1/2). (2^1)/(2^(1/2))=2^(1-1/2)=2^(1/2).
 
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  • #3
SammyS
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1. I am on the very last step of this problem, but I don't see how this equation simplifies from
2ln|(2/√(2)) + 1| - 2ln|1 + 0| to 2ln|√(2) + 1|.

2. I thought that since ln|1| = 0, then 2x0=0. And then the answer would be:
2ln|2/√(2) +1|.

I don't see how the 2 in the (2/√(2)) was canceled out to give just √(2)...

Would you please explain this to me?
Thank you very much! :D
Comment #1: Those are not equations -- there are no equal signs.


To answer your primary question:

Use the following logarithm property.
[itex]\displaystyle C\cdot\ln(u)=\ln(u^C\,) [/itex]​
 
  • #4
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...I was debating using the word "equation," but then I didn't know what else to call it... :/
Thanks, though! :)
I didn't even know that was a property. Thanks! Will remember!

So then this:

[itex]\pi[/itex][2ln|(2/√(2)) + 1|) - [itex]\pi[/itex][2ln|1|]

Shoud become this:

[itex]\pi[/itex][ln|(2/√(2)) + 1|2] - 0


From the property: C*ln(u) = ln(uC), it seems that the entire [(2/√(2)) + 1] is the u...

But that does not seem to be correct because then that squared would be:

(4/√(2)) + 3...

So I guess only the (2/√(2)) is considered the u...???

But then that would equal ln|2 + 1| = ln|3|

I don't see how the √(2) remains.

Am I saying the u equals the wrong thing?
Please help.
Thank you so much! :)
 
  • #5
Dick
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...I was debating using the word "equation," but then I didn't know what else to call it... :/
Thanks, though! :)
I didn't even know that was a property. Thanks! Will remember!

So then this:

[itex]\pi[/itex][2ln|(2/√(2)) + 1|) - [itex]\pi[/itex][2ln|1|]

Shoud become this:

[itex]\pi[/itex][ln|(2/√(2)) + 1|2] - 0


From the property: C*ln(u) = ln(uC), it seems that the entire [(2/√(2)) + 1] is the u...

But that does not seem to be correct because then that squared would be:

(4/√(2)) + 3...

So I guess only the (2/√(2)) is considered the u...???

But then that would equal ln|2 + 1| = ln|3|

I don't see how the √(2) remains.

Am I saying the u equals the wrong thing?
Please help.
Thank you so much! :)
If your question is why (2/√(2))=√(2), none of that has much to do with it. Did you miss my post?
 
  • #6
SammyS
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Did you know that [itex]\displaystyle \ \ \frac{2}{\sqrt{2}} = \sqrt{2}\ ?[/itex]

Furthermore, [itex]\displaystyle \ (\sqrt{2}\,)^2=2\ .[/itex]
 
  • #7
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Oh, wow, you guys! XD

Yes, Dick, I did miss your post! When I was scrolling through, I seemed to only see SammyS's!

Wow, √(2)/2 = √(2)!!!!! Ugh, okay! ;)

I get it now! Thanks, Dick and SammyS! :D
simple mistake! :/
 

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