# Solution of a system with exponentials

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?

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Smarter than subtracting $1$ from both sides and taking logs? –  Peter Taylor Feb 4 '12 at 12:48
yes that's the obvious way but it has a substantial bit of calculations! –  nikos Feb 4 '12 at 12:52
Not really... $\phantom{}$ –  Ｊ. Ｍ. Feb 4 '12 at 12:53

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 $N$.

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