User:Lemio/sandbox

File:Arduino100IDE

Over-Exponential functions

$2^{2\uparrow x}=e^{2^{x}*ln(2)}$

Above function on logarithmic paper gives $log(2^{2\uparrow x})=2^{x}*log(2)$

Above function on logarithmic paper gives $log(log(2^{2\uparrow x}))=$$x*log(2)+log(log(2))$

Over-exponentiation is a mathematical operation, written as $b^{n\uparrow m}$, involving three numbers, the base b, exponent (or power) n and the over-exponent (or over-power) m. When m is a positive integer, over-exponentiation corresponds to repeated exponentiation.

$b^{n\uparrow 0}=b$

$b^{n\uparrow 1}=b^{n}$

$b^{n\uparrow 2}=(b^{n})^{n}$

$b^{n\uparrow 3}=((b^{n})^{n})^{n}$

$\cdots$

just as exponentiation by a positive integer corresponds to repeated multiplication:

$b^{n}=\underbrace {b\times \cdots \times b} _{n},$

A formula to calculate $b^{n\uparrow m}$ when m is not a positive integer is:

$b^{n\uparrow m}=e^{n^{m}*ln(b)}$

Prove

$b^{n\uparrow m}=b^{n\uparrow m}$

$ln(b^{n\uparrow m})=ln(b^{n\uparrow m})$

$ln(b^{n\uparrow m})=n^{m}*ln(b)$

$b^{n\uparrow m}=e^{n^{m}*ln(b)}$

$b^{n\uparrow m}=e^{n^{m}*ln(b)}$

The inverse function

$v=b^{n\uparrow m}$

If you want to know v:

$v=e^{n^{m}*ln(b)}$

If you want to know m

$m=log_{n}(log_{b}(m))$

If you want to know n

$n=(log_{b}(v))^{\frac {1}{m}}$

If you want to know b

$b=e{\frac {ln(v)}{n^{m}}}$

$x=e^{n^{y}*ln(b)}$

$ln(x)=n^{y}*ln(b)$

${\frac {ln(x)}{ln(b)}}=n^{y}$

$y=log_{n}(log_{b}(x))$

Tests

$({10^{10^{Q-3}}})^{100^{x}}$