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!

  • $\begingroup$ use this: $$a^{b + c} = a^b * a^c $$ $\endgroup$
    – Makina
    May 4, 2021 at 12:16

2 Answers 2


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


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

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

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

$ (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]$

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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.