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Is it possible to get an explicit solution for this? 
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#1
Dec1012, 11:58 PM

P: 2

x = 10logx + 30 (log is log base 10)
I cannot get to anything other than this implicit solution. By trial and error I can tell that x must be slightly more than 1/1000 but I would like to get an exact answer. 


#2
Dec1112, 12:09 AM

HW Helper
P: 2,264

You need a function such as the Lambert W function to get an explicit solution. Real solution are approximately .0001 and 46.6925.



#3
Dec1112, 12:21 AM

P: 2

Thanks for the reply. I know you probably don't want to give me the answer right out but in looking at that Wikipedia article I still don't see how to do it. I am not able to get both x's on the same side in any manner resembling the examples from the article.



#4
Dec1112, 04:01 AM

HW Helper
P: 2,264

Is it possible to get an explicit solution for this?
It can be a little tricky. These log's are base e.
x = (10/log(10))log(x) + 30 (log(10)/10)x=log(x)+30(log(10)/10) (log(10)/10)x+log(x)=log(10^3) (log(10)/10)x+log((log(10)/10)x)log(log(10)/10)=log(10^3) (log(10)/10)x+log((log(10)/10)x)=log(log(10)/10^4) (log(10)/10)x=W(log(10)/10^4) x=(10/log(10))W(log(10)/10^4) There are log's of negative numbers in there. For finding numerical answers, you can improve your guess systematically guess 50 x = 10log10(x) + 30 guess 0 x=10^(x/103) Then in each case put the guess into the right hand side over and over until it changes very little. ie 50 10log10(50) + 30~46.9897000 10log10(46.9897000) + 30~46.7200267 ~46.6950 and so forth 


#5
Dec1112, 05:16 AM

Sci Advisor
P: 2,751

[tex]\ln(x) = ax + b[/tex] Now exponentiate both sided and put it in the form, [tex]x = e^b \, e^{ax}[/tex] Finally mult both sides by (a) and rearrange into the form, [tex] (ax) e^{ax} = k[/tex] It should then be straightforward to use the Lambert W function. 


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