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.
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 $x=\frac {\log(y)}{\log(-a)}$