I’m working on an advanced calculator featuring exponentation, tetration, hexation, pentation, roots, super roots, super logs, and etc, so far I have the first 5 finished, but I need some help with the 6th as well as the 7th, if your wondering, the code for exponents, roots, and tetration looks like this:10^(exponent)*log(base) , 10^{1/(root)}*log(base) , 10^(tetration)*log{10^(base)*log(base)} . So.. how do I find super roots? I’ve had some somewhat successful attempts, but one common error is that those attempts usually only get it right when the super root is an even number and the result is a perfect super root. I know that super root = “What tetration (super root) = (super root base)?” As for super logs, I have no idea at all how those work, and, I haven’t actually tried making them yet. So.. a little help, please?

BW_CA_24 wrote:

I’m working on an advanced calculator featuring exponentation, tetration, hexation, pentation, roots, super roots, super logs, and etc, so far I have the first 5 finished, but I need some help with the 6th as well as the 7th, if your wondering, the code for exponents, roots, and tetration looks like this:10^(exponent)*log(base) , 10^{1/(root)}*log(base) , 10^(tetration)*log{10^(base)*log(base)} . So.. how do I find super roots? I’ve had some somewhat successful attempts, but one common error is that those attempts usually only get it right when the super root is an even number and the result is a perfect super root. I know that super root = “What tetration (super root) = (super root base)?” As for super logs, I have no idea at all how those work, and, I haven’t actually tried making them yet. So.. a little help, please?

I accidentally figured out super roots (At least, I think I did.) I just did the following: tetration(tetration base),{1/(tetration base)*{2*(super root)}.

a superlog is an inverse of a tetrational function apparently

CrazyCoder1247 wrote:

a superlog is an inverse of a tetrational function apparently

I figured out super logs with ANY base..
X = Log #
Y = Base
Log(x,y) = log(x)/log(y)
Super Log(x,y) = 1+ (Log(x,y)/y)

BW_CA_24 wrote:

CrazyCoder1247 wrote:

a superlog is an inverse of a tetrational function apparently

I figured out super logs with ANY base..
X = Log #
Y = Base
Log(x,y) = log(x)/log(y)
Super Log(x,y) = 1+ (Log(x,y)/y)

Oh yeah, with tetration, the “tetration exponent”, as I call it, should be subtracted by one before actually running the operation for the ideal answer.

BW_CA_24 wrote:

BW_CA_24 wrote:

CrazyCoder1247 wrote:

a superlog is an inverse of a tetrational function apparently

I figured out super logs with ANY base..
X = Log #
Y = Base
Log(x,y) = log(x)/log(y)
Super Log(x,y) = 1+ (Log(x,y)/y)

Oh yeah, with tetration, the “tetration exponent”, as I call it, should be subtracted by one before actually running the operation for the ideal answer.

All that’s left now is to repeat this with pentation and hexation, however, I can’t find any information on the inverse of those hyperoperations, I only found one form, but it only talks about tetration, nothing else.

BW_CA_24 wrote:

BW_CA_24 wrote:

BW_CA_24 wrote:

CrazyCoder1247 wrote:

a superlog is an inverse of a tetrational function apparently

I figured out super logs with ANY base..
X = Log #
Y = Base
Log(x,y) = log(x)/log(y)
Super Log(x,y) = 1+ (Log(x,y)/y)

Oh yeah, with tetration, the “tetration exponent”, as I call it, should be subtracted by one before actually running the operation for the ideal answer.

All that’s left now is to repeat this with pentation and hexation, however, I can’t find any information on the inverse of those hyperoperations, I only found one form, but it only talks about tetration, nothing else.

One last thing, I did all of this on a mobile device, not a computer, a device with a touchscreen, which made things slightly more difficult.

Now I'm no expert but I think this should be in Advanced Topics. Reporting to be moved