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There's a constant that is very close to an integer that's referenced here:

$$e^{\pi} - \pi = 19.9990999$$

We nerds find this to be cool because it has two mysterious numbers and it's so close to being an integer.

Well, I got myself into a pickle here when I got curious how much we would need to offset $e$ and $\pi$ to make it come out to $20$. So I set myself up with a single constant $c$ in:

$$(e+c)^{\pi+c} - (\pi+c) = 20$$

And it occurs to me that I've simply been out of math too long to figure this out on my own. At the same time, I can't just 'let it go' (who does that? What the fool calls inability is what the wise man calls opportunity for growth!)

What steps would I take to solve such an equality?

Note: the tag selection didn't give me much to play with. I would prefer the tag of [Constants] but it's unavailable and alas I can't transfer my SO rep to here. :S

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I'm not sure that an exact solution can be found in terms of elementary functions. However, plenty of numerical methods exist for solving this problem. You could try Newton's method (if you want to work it out yourself) or just ask WolframAlpha. – Alex Becker Apr 9 '11 at 0:31
Another interesting think on that link, pi almost = ‎(9^2 + 19^2/22)^(1/4) – picakhu Apr 9 '11 at 1:58
up vote 3 down vote accepted

Wolfram Alpha gives a value of about 1.84105743 E-5, you can get more digits

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If you have a function $f(c)$ so that $f(0)$ almost has the value you want $y$, then a reasonable next guess is $(y-f(0))/f'(0)$. This is the first step of Newton's method.

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