I have been having some issues with simplifying the following equation:
$p\Big[ \frac{e^{ay}+e^{a(10+y)}}{e^{a(10-y)}+e^{a(20-y)}} \Big]=1-p$
where $y$ is the variable, $a$ is a parameter and $p$ is a constant.
I should be getting an expression in terms of $y$, but I do not know how to proceed. I tried the following:
$\Big[ \frac{e^{ay}+e^{a(10+y)}}{e^{a(10-y)}+e^{a(20-y)}} \Big]=\frac{1-p}{p}$
$ln\Big[ \frac{e^{ay}+e^{a(10+y)}}{e^{a(10-y)}+e^{a(20-y)}} \Big]=ln\Big[\frac{1-p}{p}\Big]$
I do not know how to simplify the expression in brackets. Thanks in advance!
Best Answer
$$\frac{e^{ay}+e^{a(10+y)}}{e^{a(10-y)}+e^{a(20-y)}}=\frac{e^{ay}(1+e^{10a})}{e^{-ay}(e^{10a}+e^{20a})}$$ Now you can easily get $y$ in one expression. Now lets look at the other part: $$\frac{1+e^{10a}}{e^{10a}+e^{20a}}$$ $$\frac{1+e^{10a}}{e^{10a}(1+e^{10a})}=e^{-10a}$$