Is it possible to get an explicit solution for this?

pondering

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.

lurflurf

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

pondering

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.

lurflurf

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/10-3)

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

uart

First convert your log to base "e", and then write your equation in the form,

[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.