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I have the following

$(1-a^x)/x=b$

Can this be solved for x ? (if yes, how, if not why)

I have gotten to many forms, but can't seem to isolate x.

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  • $\begingroup$ Post some of the forms you found. $\endgroup$ – Aaron Maroja Jun 8 '15 at 14:52
  • $\begingroup$ In general an equation can´t be solved algebraically for x, if x appears as a factor and as an exponent. $\endgroup$ – callculus Jun 8 '15 at 14:59
  • $\begingroup$ @AaronMaroja will post tomorrow morning as the calculus in on a colleagues paper (but it's mostly high-school stuff) $\endgroup$ – norb Jun 8 '15 at 15:36
  • $\begingroup$ @calculus I got to this statement myself but was not satisfying as no demonstration was included (by the answer provider). this is why I stated the question like I did... $\endgroup$ – norb Jun 8 '15 at 15:37
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\begin{align} (1-a^x)/x&=b\quad (1) \end{align}

Assuming $x\ne 0,b\ne0,a>0,a\ne1$, let $y=a^x\ (x=\ln(y)/\ln(a))$, then we can rewrite (1) as \begin{align} \frac{\ln(y)}{1-y}&= \frac{\ln(a)}{b}. \quad (2) \end{align}

Since $\frac{\ln(y)}{1-y}<0$ for $y>0,y\ne 1$, real solution for $y$ exists, if rhs of (2) is also negative, that is, either $a\in(0,1),b>0$ or $a>1,b<0$. Let $\frac{\ln(a)}{b}=w<0$: \begin{align} \frac{\ln(y)}{1-y}&= w<0. \quad (3) \end{align}

Solution of (3) also splits into two cases, $|w|<1$ and $|w|>1$: \begin{align} y&=\begin{cases} \operatorname{W}_{-1}(w\exp(w))/w,& |w|<1 \\ \operatorname{W}_{0}(w\exp(w))/w,&|w|>1 \end{cases} \end{align} where $\operatorname{W}_{0},\operatorname{W}_{-1}$ are two real branches of the Lambert W function.

Finally, solution of (1) is: \begin{align} x&=\begin{cases} \displaystyle\frac{\ln\left(\operatorname{W}_{-1}(w\exp(w))/w\right)}{\ln(a)},& |w|<1 \\ \displaystyle\frac{\ln\left(\operatorname{W}_{0}(w\exp(w))/w\right)}{\ln(a)},& |w|>1 \end{cases} \end{align}

Example:

$$ \begin{array}{rrrrrr} \hline a & b & w=\ln(a)/b & y & x & (1-a^x)/x \\ \hline 0.01 & 1.0 & -4.60517 & 0.01050 & 0.98950 & 1.0 \\ 0.80 & 15.0 & -0.01488 & 404.50727 & -26.90048 & 15.0 \\ 3.00 & -0.2 & -5.49306 & 0.00421 & -4.97894 & -0.2 \\ 4.00 & -0.2 & -6.93147 & 0.00098 & -4.99508 & -0.2 \\ 7.00 & -5.0 & -0.38918 & 5.27112 & 0.85422 & -5.0 \end{array} $$

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I assume that $a,b>0$ and $a\neq 1$.

First rewrite the equation as $$ \frac{1-\mathrm e^{x\ln a}}b=\frac{x \ln a}{\ln a}$$ and let us write $y=\frac{\ln a}b- x\ln a$. We get $y=\frac{\ln a}b\mathrm e^{\frac{\ln a}b-y}$ or $$ya^y=\frac{\ln a}ba^{1/b}.$$ The solution of this equation is given, if $\boxed{\frac{\ln a}ba^{1/b}<-\frac1{\mathrm e}}$, by the Lambert W function : $y=W\left(\frac{\ln a}{b}a^{1/b}\right)$ or $$x=\frac{1}b-\frac1{\ln a}W\left(\frac{\ln a}{b}a^{1/b}\right).$$

EDIT The same calculation actually works for $b<0$ with $a>1$. The framed condition remains identical.

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  • $\begingroup$ I found the Lambert-W function but had no idea what to do with it (personal knowledge constraints)... will come back tomorrow (evening here) with a proper comment $\endgroup$ – norb Jun 8 '15 at 15:38

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