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I've found this interesting equation on the web: $$p-1 = (1 - e^{\alpha-\beta t})^{t+1}$$

which has to be solved for t, considering that the parameters: $\alpha, \beta, p$ are defined correctly.

First thought was to use some logarithms: $$\ln(p-1)=(t+1)\cdot ln(1-e^{\alpha-\beta t})$$ $$t+1=\frac{ln(p-1)}{ln(1-e^{\alpha-\beta t})}$$

Or, alternatively, I have: $$t+1=log_{1-e^{\alpha-\beta t}}(p-1)$$

I also thought about some substitutions than can be made, or an analytical approach.
Any guidance will be greatly appreciated.

Alexandru Dinu
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    Depending on what values the parameters can take, maybe you can potentially replace $1-e^{a-bt}$ with a good approximation, maybe linear? – Sandeep Silwal Jun 11 '14 at 20:23
  • If I let $\alpha = 0, \ \beta = -1$, then I define $f(x) = log_{1-e^x}(p-1)- x-1$, I can easily plot this and see for which values of p I get a convenient, solvable $f(x) = 0$. I want to know if I can write t as a standalone expression of $\alpha, \ \beta, \ p$. – Alexandru Dinu Jun 11 '14 at 20:28

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