- #1

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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.

- Thread starter pondering
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- #1

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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

lurflurf

Homework Helper

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- #4

lurflurf

Homework Helper

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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

- #5

uart

Science Advisor

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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.

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