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As title, what's the approximation of $\ln(1-x)$ in terms of only $\ln(x)$, its linear combination (i.e. $a\ln(x)+b$) or $(\ln(x))^2$ ,where $x$ is between 0 and 1?

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    I would guess that a nonlinear regression should give you good parameter estimates for $a$ and $b$. – 2017-02-11
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    I'm looking for an approximation better than this. A good approximation I have so far is $\ln(x)-\ln(\tan(\pi x/2))$ yet this violates the required conditions – 2017-02-11
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    Why not simply use a polynomial? – 2017-02-11
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    This is actually part of my research.Long story short. I have a function that looks like this: $\ln(V_i)=\ln(P_i)+\ln(1-(C_i k)/(P_i))+c$. Now I'd like to find $\ln(V_2/V_1)$ in terms of $\ln(P_2/P_1)$ and $\ln(C_2/C_1)$ – 2017-02-11
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    $P$ and $C$ are some price and cost indices so only their rate of change (i.e. log difference) can be applied to calculate $V$. I'm now trying to approximate $\ln(1−(C_i k)/(P_i))$. – 2017-02-11
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    What is $k$? Is it constant? Or also a parameter depending on $i$? – 2017-02-11
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    yep, a constant not depending on $i$. – 2017-02-11
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    I appreciate (really much!) the fact you got rid of unnecessary constants, but there is no way such approximation can be accurate: $\log(x)$ and $\log(1-x)$ have singularities at different points. – 2017-02-11

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This is not an analytical answer to your question.

I used a nonlinear Regression with the least squares error measure. For $\ln(1-x)$ approximated by $a\ln(x)+b$ I received $a\approx 0.19838137...$ and $b \approx 0$. But the fit is very poor.

The problem is partially caused as you are trying to approximate a decreasing function by an increasing function. Even a linear approximation would work better.