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I want to solve the following system:

$1 + (0.2\pi/W)^{2N} = (1/0.89125)^2$

and

$ 1 + (0.3\pi/W)^{2N} = (1/0.17783)^2 $

but i can't see how i can do that without getting to many confusing calculations. Is there a 'smarter' way to solve this system?

  • 0
    Not really... $\phantom{}$2012-02-04

2 Answers 2

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Let's denote :

$A =(1/0.89125)^2$ , and $B=(1/0.17783)^2$ . So:

$(0.2\pi/W)^{2N}=A-1$

$(0.3\pi/W)^{2N}=B-1$

Now divide first equation by second :

$\left(\frac{2}{3}\right)^{2N}=\frac{A-1}{B-1} \Rightarrow N = \frac{1}{2} \cdot \log_{2/3} \frac{A-1}{B-1}$

After you find variable $N$ substitute it into one of the starting equations .

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The first equation can be rearranged like so:

$2N\log(0.2\pi/W) = \log((1/0.89125)^2-1)$

and a similar thing can be done for the second equation. Perform a division to cancel out the $N$, leading to

$\frac{\log(0.2\pi/W)}{\log(0.3\pi/W)} = \frac{\log((1/0.89125)^2-1)}{\log((1/0.17783)^2-1)}$

or

$\frac{\log(0.2\pi)-\log\,W}{\log(0.3\pi)-\log\,W} = \frac{\log((1/0.89125)^2-1)}{\log((1/0.17783)^2-1)}$

Solving for $\log\,W$, and then $W$, should be a snap. Once you have the value of $W$, substitute into any of the two original equations, or the equation like the first one I gave in this answer to solve for $N$.