Equation solving

**bjkrishna** · January 5th 2010, 05:19 PM

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

Please help me to solve the equation

Thanks

**evouga** · January 5th 2010, 11:34 PM

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).

**abender** · January 6th 2010, 01:04 AM

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

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