Thread: Equation solving

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

Please help me to solve the equation


Thanks

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).

Originally Posted by bjkrishna

Find the value of $\displaystyle P_1$ in the equation

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

Wow, this is quite a mess. Alright, seeing as I can't sleep, I might as well give this a go here.


$\displaystyle
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
$

$\displaystyle \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
$

$\displaystyle \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
$


$\displaystyle \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
$


$\displaystyle \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
$


$\displaystyle \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
$


$\displaystyle \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