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I suppose we have to solve the given non-linear logarithmic equation.

solving the given equation:

Log\frac{0.9}{x}=4.15\ Log\ \frac{1.1}{1-x}

gives  x = 0.218198  approximately.

Log\frac{0.9}{x}=4.15\ Log\ \frac{2(1-0.45)}{1-x}\\\\Log\ 0.9-Log(x)=4.15*Log\frac{1.1}{1-x}\\\\log\ 0.9-log(x)=4.15*Log1.1-4.15*log(1-x)\\\\-0.2175 -log(x)=-4.15*log(1-x)\\\\let\ x=y^{4.15}\\\\-0.2175-4.15log(y)=-4.15\ log(1-y^{4.15})\\\\log(1-y^{4.15})-log(y)=0.05242\\\\log\frac{1-y^{4.15}}{y}=0.5242\\\\\frac{1-y^{4.15}}{y}=10^{5242}=1.12828\\\\y^{4.15}+1.12828y=1

This equation can be recursively used to find y.  After finding y to the required precision, we find  x = y^4.15..

for example:  
y = 0.1    =>  y =

1 5 1