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Math Help - Equation solving

  1. #1
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    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

    Please help me to solve the equation


    Thanks
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  2. #2
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    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).
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  3. #3
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    Quote Originally Posted by bjkrishna View Post
    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.


    <br />
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<br />

     \implies<br />
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<br />

     \implies<br />
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<br />


     \implies<br />
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<br />


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


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


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


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

    -Andy
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