May I ask why you use commas for decimal points?. Is that one of them wacky European things?.
\(\displaystyle .4005=.8\left(ln(\frac{.4}{x})+2.25ln(\frac{1-x}{.6})\right)\)
\(\displaystyle .500625=\left(ln(\frac{.4}{x})+2.25ln(\frac{1-x}{.6})\right)\)
Expand using log laws:
\(\displaystyle -.267558=ln(x)-2.25ln(1-x)\)
\(\displaystyle e^{-.267558}=\frac{x}{(1-x)^{\frac{9}{4}}}=(x^{\frac{-4}{9}}(1-x))^{\frac{-9}{4}}\)
\(\displaystyle x^{\frac{-4}{9}}-x^{\frac{5}{9}}-1.1262738=0\).............
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Let \(\displaystyle u=x^{\frac{1}{9}}\). Which leads to:
\(\displaystyle u^{9}+1.1262738u-1=0\)
Factoring is really rough here, but can be done with tech. This has 9 solutions, but only 1 is real.
By factoring, the real root is:
\(\displaystyle .8812782-u=0\)
\(\displaystyle x^{\frac{1}{9}}=.8812782\)
\(\displaystyle x=(.8812782)^{9}=\fbox{.32064}\)
Of course, the accuracy can be to however many decimal places you want.
Another way is to use Newton's Method to solve this...assuming you are familiar with a little calculus. Frankly, that would be my method of choice starting at
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