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Assume we have a simple equation

  • $a^x = y, \quad a, x \in \mathbb{R}, \; a \neq 0$

from where $x$ needs to be evaluated. If we set a restriction $a > 0$, there is a simple logarithm expression available

  • $x = \log_a y = \frac{\log y}{\log a}$.

Still I'm not sure how to deal with cases such as $(-2)^x=-8$.
Eventually, I would like to solve $\theta$ from

  • $a^{\theta} \exp (-\theta \sum\limits_{k=1}^{N} f(k)) = c \qquad a,c \neq 0, \; N \in \mathbb{N}^{+}$

Any help would be highly appreciated.

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If $a < 0$ and $y < 0$, then $x$ is restricted to odd positive integers. –  rgettman Jun 25 '13 at 20:21

1 Answer 1

up vote 0 down vote accepted

If $a,y \lt 0$ in $\Bbb R$ we are restricted to $x$ being an odd integer and we can use $x=\frac {\log(-y)}{\log(-a)}$. If $a \lt 0, y \gt 0$ we are restricted to $x$ being an even integer and can use $x=\frac {\log(y)}{\log(-a)}$

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Thank you for your help. So basically the analysis will be divided into several conditions. –  Tim Jun 25 '13 at 20:43
    
@Tim: basically you need the argument of $\log$ to be positive. –  Ross Millikan Jun 25 '13 at 20:54

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