How to simplify sum of exponential functions

Question

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!

Answer 1

$$\frac{e^{ay}+e^{a(10+y)}}{e^{a(10-y)}+e^{a(20-y)}}=\frac{e^{ay}(1+e^{10a})}{e^{-ay}(e^{10a}+e^{20a})}$$ Now you can easily get $y$ in one expression. Now lets look at the other part: $$\frac{1+e^{10a}}{e^{10a}+e^{20a}}$$ $$\frac{1+e^{10a}}{e^{10a}(1+e^{10a})}=e^{-10a}$$

Answer 2

$\Big[ \frac{e^{ay}+e^{a(10+y)}}{e^{a(10-y)}+e^{a(20-y)}} \Big]*\Big[\frac{e^{ay}}{e^{ay}} \Big]=\Big[\frac{1-p}{p}\Big]$

$\Big[ \frac{(e^{2ay}+e^{10+2ay)}}{e^{10a}+e^{20a)}} \Big]=\Big[\frac{1-p}{p}\Big]$

$\Big[ \frac{e^{2ay}+e^{(10+2ay)}}{e^{10a}+e^{20a}} \Big]=\Big[\frac{1-p}{p}\Big]$

$ (e^{2ay})\Big[ \frac{1+e^{10a}}{e^{10a}+e^{20a}} \Big]=\Big[\frac{1-p}{p}\Big]$

$ (e^{2ay})=\Big[ \frac{e^{10a}+e^{20a}}{1+e^{10a}} \Big]\Big[\frac{1-p}{p}\Big]$

$2ay=\ln({\Big[ \frac{e^{10a}+e^{20a}}{1+e^{10a}} \Big]\Big[\frac{1-p}{p}\Big]})$

$y=\frac{1}{2a}\ln({\Big[ \frac{e^{10a}+e^{20a}}{1+e^{10a}} \Big]\Big[\frac{1-p}{p}\Big]})$

$y=\frac{1}{2a}\ln({\Big[ \frac{1}{e^{10a}} \Big]\Big[\frac{1-p}{p}\Big]})$