Math problem

quote:

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skim
Also, does anyone know any good math forum?

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-UltimaX-

"You wished for a white christmas... Now go shovel your wishes!"

I started with

100n^2 < 2^n

ln(100) + ln(n^2) < n*ln(2)
                                   
4.6 + 2*ln(n) 
------------- < n
   ln(2)
                                
6.67 + 2*ln(n) < n

  

EDIT: scratch what I did above, its wrong. I started solving it, but I think at some point you'll have to use substitution to figure it out. I think it might get a bit lengthy, but if you're lucky someone here will probably know the best method and can help you better.

[edited by - nervo on August 28, 2003 11:53:34 PM]

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Original post by Anonymous Poster

quote:

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Original post by skim
How do you solve for n: 100*n^2 < 2^n
I know how to solve it graphically but how do you solve for n algebraically?

Also, does anyone know any good math forum?

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100 * n^2 < 2^n

log 100 + 2 < n log 2
n n

n > log 98 + 2 / log 2
n n

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What the hell is that?

Can someone please confirm whether this equation is able to be solved algebraically or not. Its killing me trying to solve it through algebra. I''m beginning to think this is one of those ones that can only be solved through trial and error or graphically, somewhat like the folium of descartes. Its killing meh!

It can not be algebraically solved.

You can only solve by numerical methods. I could not remember
now the what methods exists but any numeric book contains such information.

I am leaning towards my feeling that it is impossible to solve algebraically. The expression looks quite simple, but the key is that it is mixing an exponential function with a common algebraic fixed power expression. This is not a type of problem that I have come across with a mixture of this kind up through calculus.

I found this quote on a college website that indicates expressions of this sort:

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Sometimes, it is impossible to solve an equation involving logarithms or exponential functions algebraically. This is especially true when the equation involves transcendental (logs and/or exponentials) and algebraic components

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[edited by - nervo on August 29, 2003 2:15:39 AM]

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Original post by beehorf
It can not be algebraically solved.

You can only solve by numerical methods. I could not remember
now the what methods exists but any numeric book contains such information.

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Thanks, you confirmed by suspicions. The quote above also helped that also. I'm beyond the introductory calculus level and if this was a problem that could have been solved through algebra and I still could not have found it, I would be hanging my head in deep shame right now.

[edited by - nervo on August 29, 2003 2:16:07 AM]

what''s wrong is anonymous poster''s reply.
What would you want as a result?
A number expressed in common (usual) functions is thge best you can get!

I realize that n cannot be solved with a basic numeric value algebraically, but how in the world did he/she get:

"log 100 + 2"

How did this 2 just popup and where did the n variable go?

and what is "n n"?

EDIT:

This is how I would isolate n:

100n^2 < 2^n
       
log(100) + log(n^2) < log(2^n)
        
log(100) + 2log(n) < nlog(2)
       
log(100) + 2log(n) 
------------------ < n
     log(2)

then doing the trivial arithmetic to get numeric values for log(2) and log(100)

[edited by - nervo on August 29, 2003 6:23:04 AM]

Thank you all for your quick replies. I appreciate all your help. I think I will look into numeric methods to see how I can solve this problem if I ever encounter it again.

The second question remain unanswered. Does everyone know any good math forums on the internet?

Check the Forum FAQ (in this very forum)

EDIT: Link because I felt nice

[edited by - Cedric on August 29, 2003 11:05:52 AM]