## Is it possible to get an explicit solution for this?

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

Recognitions:
Homework Help

## 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.

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

Recognitions:
 Quote by 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.
First convert your log to base "e", and then write your equation in the form,

$$\ln(x) = ax + b$$

Now exponentiate both sided and put it in the form,

$$x = e^b \, e^{ax}$$

Finally mult both sides by (-a) and rearrange into the form,

$$(-ax) e^{-ax} = k$$

It should then be straightforward to use the Lambert W function.