1. ## Equation solving

I need to find the value of $P_1$ in the equation
$2^{\frac{B_1}{B_2}log_2(1+\frac{h^2P_1}{N_1})-R}+\frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1$

Thanks

2. Is it known that this equation can be solved analytically? If not, I strongly suspect you'll need to resort to plugging in the other constant and solving for the root numerically (eg using Newton's Method).

3. Originally Posted by bjkrishna
Find the value of $P_1$ in the equation

$2^{\frac{B_1}{B_2}log_2(1+\frac{h^2P_1}{N_1})-R}+\frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1$
Wow, this is quite a mess. Alright, seeing as I can't sleep, I might as well give this a go here.

$
2^{\frac{B_1}{B_2}log_2(1+\frac{h^2P_1}{N_1})-R}+\frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1
$

$\implies
2^{\frac{B_1}{B_2}log_2(1+\frac{h^2P_1}{N_1})} \bigg{(}2^{-R} \bigg{)}+\frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1
$

$\implies
2^{\frac{B_1}{B_2}log_2(\frac{N_1+h^2P_1}{N_1})} \bigg{(}2^{-R} \bigg{)}+\frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1
$

$\implies
2^{\frac{B_1}{B_2} \big{(}log_2(N_1+h^2P_1) - log_2(N_1) \big{)}} \bigg{(}2^{-R} \bigg{)}+\frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1
$

$\implies
\bigg{[} 2^{\frac{B_1}{B_2} \big{(}log_2(N_1+h^2P_1)\big{)}}\bigg{]}
\bigg{[} 2^{-\frac{B_1}{B_2} \big{(}log_2(N_1) \big{)}}\bigg{]}
\bigg{[} 2^{-R} \bigg{]}
+ \frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1
$

$\implies
\bigg{[} 2^{\big{(}log_2(N_1+h^2P_1)\big{)}\big{(}\frac{B_1 }{B_2}\big{)}}\bigg{]}
\bigg{[} 2^{\big{(}log_2(N_1) \big{)}\big{(}-\frac{B_1}{B_2}\big{)}}\bigg{]}
\bigg{[} 2^{-R} \bigg{]}
+ \frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1
$

$\implies
\big{(}(N_1 + h^2 P_1)^{\frac{B_1}{B_2}} \big{)}
\big{(} N_1^{-\frac{B_1}{B_2}} \big{)}
\big{(} 2^{-R} \big{)}
+ \frac{B_1h^2(P-P_1)}{B_2(N_1+h^2P_1)}=1
$

I entered the rest into a program because I like being sane.... and it told me no solution.... FWIW

-Andy