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I've reduced a set of equations to $V_t\ln(\frac{I_c}{I_s}) = V_{cc} - I_c*1000$

I need to solve this for $I_c$; everything other than $I_c$ are constants. I don't even know where to begin; normally I would exponentiate both sides but then I have an exponent of $I_c$ and I've just moved my problem without actually solving it ...

So I suppose the more general question is: How can I solve an equation with both $ln(x)$ and $x$ in the equation?

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  • $\begingroup$ The equation cannot be solved algebraically for $I_c$. $\endgroup$ – callculus Oct 19 '15 at 4:24
  • $\begingroup$ Well! I guess my prof isn't likely to get mad at me for using wolfram alpha then, but how would(could) it be solved? $\endgroup$ – Daniel B. Oct 19 '15 at 4:26
  • $\begingroup$ I don´t know, algebraically it is not possible. $\endgroup$ – callculus Oct 19 '15 at 4:41
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There is no analytical solution in terms of elementary functions.

However, the solution can be expressed in terms of Lambert function (which is such that $x=W(x)\, e^{W(x)}$); applied to your case, this will lead to $$I_c=\frac {V_t}{1000} W(z)$$ using $$z=\frac {1000\,I_s}{V_t}\,e^{\frac{V_{cc}}{V_t}}$$

In the Wikipedia page, you will find nice approximations of this function.

In fact, any equation which can write $A+Bx+C\log(D+Ex)=0$ has solutions which can be expressed in terms of Lambert function.

Otherwise, only numerical methods such as Newton would solve the problem.

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