General Formula for all arithmetic operations

[azure crane]

${}^{k}a=log_{a,(x-k)}({}^xa)$

[dawn finch]

RoyalBanana

[azure crane]

Fair

[jovial rock]

Do you remember?

Can we solve this (e^^2)^^x with the new rules?

The third one?

[azure crane]

? What new rules

[jovial rock]

This

I think log(e^^2)(-x)(1)

[azure crane]

Wdym

[jovial rock]

Plugin in 1 for x, to cancel out the super log and the e^^2

Like the thing you just send in

It was the Same thing just with x = 1

[azure crane]

No I mean wdym by this

Idk which input is which

bc of the parenthesis

$log_{({}^2e,-x)}(1)$?

[dawn finch]

RoyalBanana

[jovial rock]

Yeah

Like that exactly

Hmm

Is this possible: (e^^2)^^2 = e^^x ?

[jovial rock]

If we could put the (e^^2)^^x into one tetration we could take the natrual super log

[azure crane]

ye

[jovial rock]

This looks Pretty good but we have this e^^2

[azure crane]

$log_{{}^2e}(e)=?$

[dawn finch]

RoyalBanana

[azure crane]

could help

[jovial rock]

Yeah true

[azure crane]

Yooo

$log_{{}^2x}(x)=\frac{1}{x}$

[dawn finch]

RoyalBanana

[jovial rock]

Really?

[azure crane]

(saw it from the graph)

[jovial rock]

This would be epic

[azure crane]

(the points at 0 and before are just bc it can't calculate 0 and negative bases etc)

[jovial rock]

Damn

[azure crane]

Ok

interesting

btw a useful thing to know is $\log_{a,-b}(x)$ is the inverse of $\log_{a,b}(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah true

[jovial rock]

We need the second parameter as 1 and the 3rd parameter e

[azure crane]

oh that makes sense (x^x)^(1/x)=x^(x*1/x)=x^1=x

[jovial rock]

Then it‘s 1/e

[jovial rock]

Oh nice

[azure crane]

$log_{({}^ax)}(x)=\frac{1}{{}^{(a-1)}x}$

[dawn finch]

RoyalBanana

[jovial rock]

Nice

log(e^^2)(log(e^^2)s(e) -k)(e)

What does k need to be so the iteration is 1?

[azure crane]

$log_{({}^2e),(log^s_{({}^2e)}(e) -k)}(e)$

[dawn finch]

RoyalBanana

[jovial rock]

Wait

[azure crane]

$=log_{({}^2e),(log^s_{({}^2e)}(e) -k-1)}(1/e)$

[jovial rock]

So this is (e^^2)^^2 ?

[dawn finch]

RoyalBanana

[azure crane]

Wow $log_{({}^2x)}(\frac{1}{x})=-\frac{1}{x}$

[dawn finch]

RoyalBanana

[jovial rock]

Uh

[azure crane]

yes

it works

[jovial rock]

Can you write it in words, it doesn‘t show me the pic

[azure crane]

log(x^^2)(1/x)=-1/x

[jovial rock]

Oh

Thats special

log(e^^2)(log(e^^2)s(e) -k)(e), so (e^^2)^^(log(e^^2)s(e) -1) = 1/e

I replaced k with log(e^^2)s(e) -1 so it is only 1 iteration

And we know that with one iteration log(e^^2)(1)(e) = 1/e

So (e^^2)^^(log(e^^2)s(e)-1) = e^^lns(1/e)

Not sure if this is correct but would be epic

[azure crane]

Can you explain this

I'm lost

[jovial rock]

So first, we know that log(e^^2)(1)(e) = 1/e right?

Wait a sec

Yeah right

Do you understand How I got this? log(e^^2)(log(e^^2)s(e) -k)(e)

[azure crane]

yeah

[jovial rock]

This thing

[azure crane]

I forgor

[jovial rock]

This is what we had for (e^^2)^^x

So lets replace the x with k

[azure crane]

Ah you used #1330832852819906661 message

[jovial rock]

Now it‘s (e^^2)^^k = log(e^^2)(log(e^^2)s(x)-k)(x)

Yeah

[azure crane]

yeah

Ok

[jovial rock]

Is this clear?

[azure crane]

Sure

[jovial rock]

Now we want to have this iteration value equal to 1 so we can take the solution 1/e

Right?

[azure crane]

Can you keep the two sides of the equation together

so no forgetting

[jovial rock]

Yeah, I do

[azure crane]

So (e^^2)^^k = log(e^^2)(log(e^^2)s(e)-k)(e)

[jovial rock]

What is log(e^^2)s(x)-k = 1

[jovial rock]

Yeah

[azure crane]

k=log(e^^2)s(x)-1

[jovial rock]

Right

Now you have it

Put this value in for k on the other side

And you have equal to 1/e

[azure crane]

So (e^^2)^^(log(e^^2)s(e)) = log(e^^2)(1)(e)

[jovial rock]

Yep

[azure crane]

the ones on the left cancel

[jovial rock]

Yes

[azure crane]

e=log(e^^2)(1)(e)

[jovial rock]

1/e

[azure crane]

?

[jovial rock]

.

[azure crane]

That is log not the superlog

[jovial rock]

Ye I know

[azure crane]

so how are you getting 1/e

[jovial rock]

The Right side is normal log

[azure crane]

Um one of us made a mistake somewhere bc this is saying e=1/e

[jovial rock]

Ye I See

You forgot the -1

[jovial rock]

For the k

It‘s log(e^^2)s(e)-1

[azure crane]

Ooh then it would be log(...)(0)(e)=e that's why it was wrong

[jovial rock]

Yep

[azure crane]

No reason it has to be just 1

[jovial rock]

It has

Ohh

Wait

Now i see

[azure crane]

${}^{(log^s_{({}^2e)}(e)-a)}({}^2e)= log_{({}^2e),a}(e)$

[dawn finch]

RoyalBanana

[azure crane]

alr

My brain hurts

[jovial rock]

No problem

[jovial rock]

But you see this right?

[azure crane]

btw this is just the original formula with a=e^^2, x=e, and k=a

[jovial rock]

Damn

The x is a bit diffrent

[azure crane]

?

[jovial rock]

Oh nvm

Sry

[azure crane]

So anyway then ${}^{(log^s_{({}^2e)}(e)-1)}({}^2e)=\frac{1}{e}$

[jovial rock]

But we were able to simplify this: (e^^2)^^(log(e^^2)s(e)-1) = e^^lns(1/e)

[dawn finch]

RoyalBanana

[jovial rock]

Yeah

[azure crane]

how'd you get e^^lns(1/e)?

ooh

yeah duh

[jovial rock]

What would happen if we increase the height of e^^x

[jovial rock]

Yep

I did it like that so we can see how a tetration tower got simplified

[azure crane]

Wdym

[jovial rock]

For example (e^^3)^^x instead of (e^^2)^^x

And find a formula for that

[azure crane]

Btw Therefore.. $log^s_{({}^2e)}(e)=log^s_{({}^2e)}(\frac{1}{e})+1$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah true

[azure crane]

ah duh bc $log_{({}^2e)}(e)=\frac{1}{e}$

[dawn finch]

RoyalBanana

[jovial rock]

Wait

What is log(e^^3)(e)?

[azure crane]

^

$\frac{1}{e^e}$

[dawn finch]

RoyalBanana

[azure crane]

Btw if it's useful $\log_a(x)=\frac{ln(x)}{ln(a)}$

[dawn finch]

RoyalBanana

[jovial rock]

(e^^3)^^(log(e^^3)s(e)-1) = e^^lns(1/e^e)

(e^^x)^^(log(e^^x)s(e)-1) = e^^lns(1/e^^(x-1))

nice

[azure crane]

What was the goal

[jovial rock]

Simplyfing (e^^x)^^x

Just bring x into the form: log(e^^x)s(e)-1

uh

The upper x

The second one

Now we Can solve stuff like (e^^x)^^x = 2 I guess

And the third equation

[azure crane]

do it

[jovial rock]

Uh

Soo

[azure crane]

lol

[jovial rock]

lns((e^^2)^^x)

lemme do x = 2

Wait

e^^x = (e^^2)^^2

huh

wait

[azure crane]

${}^{(log^s_{({}^ae)}(e)-1)}({}^ae)=\frac{1}{{}^{(a-1)}e}$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah, like that

[jovial rock]

log(e^^2)s(e^^x) = 2

I‘m slow sry

[azure crane]

nah youre good!

[jovial rock]

(log(e^^2)s(e^^x) * e)/2) = e

I think

[jovial rock]

Thanks

log(e^^2)s((log(e^^2)s(e^^x) * e)/2)) = log(e^^2)s(e)

log(e^^2)s((log(e^^2)s(e^^x) * e)/2))-1 = log(e^^2)s(e)-1

(e^^2)^^(log(e^^2)s((log(e^^2)s(e^^x) * e)/2))-1 )= (e^^2)^^(log(e^^2)s(e)-1)

I think thats right

(e^^2)^^(log(e^^2)s((log(e^^2)s(e^^x) * e)/2))-1 )= 1/e

Help

Big equation

Now solve for x?

Or just replace x with lns((e^^2)^^2)

[azure crane]

h

[jovial rock]

Not sure but I think it‘s this

There is a iterated super log

[azure crane]

Did you mean x instead of the 2s

[jovial rock]

Uhh

Where?

Oh I forgot the s for the second super log

[azure crane]

it says e^^2 twice

[jovial rock]

It‘s 3 times here

[azure crane]

twos not "the two s's" lol

[jovial rock]

Hmm

[azure crane]

Oh is this the equation you're solving

[jovial rock]

Ye

[azure crane]

Just take log(e^^2)s(both sides)

[jovial rock]

Oh right

[azure crane]

^

$log_{({}^2e)}(\frac{1}{e})=-\frac{1}{e}$

[dawn finch]

RoyalBanana

[jovial rock]

Whoa

This is the farthest I have gotten now

I replaced x with lns((e^^2)^^2)

Because it nicely cancel out with e^^x

And then it‘s 1 + log(e^^2)s(2)

[azure crane]

Oh my god

Wow

I just realized something

this is probably an obvious log fact ofc

but I didn't know it

$\log_a(x)=\frac{1}{log_x(a)}$

[dawn finch]

RoyalBanana

[jovial rock]

Oh really?

Does that work?

[azure crane]

^

[jovial rock]

Oh I see

[azure crane]

So with this you can easily swap the annoying e^^x out of the base

[azure crane]

Actually this is quite obvious bc log(1/x)=-log(x) lol

Is it just me or does it feel like there has to be some nice solution to

$\ln^s(a)=ln_x(a), x=?$

[dawn finch]

RoyalBanana

[azure crane]

Wait what

This wasn't supposed to be easy

$x=a-\ln^s(a)$

[dawn finch]

RoyalBanana

[azure crane]

I did some proving in dms one sec

Wait wait so $\ln^s(a)=\ln_{a-\ln^s(a)}(a)$??

[dawn finch]

RoyalBanana

[jovial rock]

Huh

This looks like it‘s true

[azure crane]

Wait

no

I made a mistake

ah

[jovial rock]

Oh

So I was solving now all the time my stupid equation and I got this

Oh

The Last Part is wrong

[azure crane]

When did the x get replaced with 2

Wait

What's the part at the top

?

Ohhhh

[jovial rock]

Do you undestand?

[azure crane]

Using this?

[jovial rock]

I only used it for this

[azure crane]

I think you made a mistake

somewhere

[jovial rock]

Where?

At the end I know

[azure crane]

idk how you got x there

and how'd you get *e

[jovial rock]

x = lns((e^^2)^^2)

[azure crane]

What are you inputting that into

[jovial rock]

I started with this here

I solved for 1/e

And then I replaced the value in log(e^^2)s(e^^x)

lns cancels out because of the e

[azure crane]

That's not what I'm asking

where did you get that top equation from

[jovial rock]

What do you mean?

x = lns(…) ?

[azure crane]

Like *e?

wheres that coming from

[azure crane]

no the equation below that

[jovial rock]

Ah

[jovial rock]

Here

[azure crane]

How would that even get in there

[jovial rock]

And the messages above this one

[azure crane]

Um where did the *e come from here?

Ooh

I see

you just multiplied by e/2

[jovial rock]

(log(e^^2)s(e^^x) )/2) = 2, (log(e^^2)s(e^^x) )/2) = 1 and then both sides * e

Yeah

[azure crane]

You forgot the /2

[jovial rock]

Shit

[azure crane]

yeah

That's why you gotta triple check with these kinds of things lol

[jovial rock]

True haha

I’ll try again

Now

I think that’s right

What is this level of math

I think using these new operators and rules you can solve every algebraic problem

[azure crane]

Can you triple check

[jovial rock]

Okay

[jovial rock]

What if I did x = 3, would the equation solve for 3 like it did for 1?

I’m pretty sure it’s true

But not 100%

[azure crane]

uhh

[jovial rock]

I checked it

[jovial rock]

Not this one btw

[azure crane]

Can you rewrite all the steps in a clean order

all in one place

so it's easy to see

This is cool

[jovial rock]

I’ll do tomorrow, it’s 12:26 AM for me and I need to go to sleep

Maybe I’ll find a general formula for x then

And I’ll try to approx the result to check if it’s true

[azure crane]

alr

[jovial rock]

Wait

It’s actually not very true

It’s log(e^^2) not lns

So the 2 e’s with the tetration are actually (e^^2)^^

A small problem but it doesn’t change anything with 2 luckily

This equation would allow us to express all numbers in form of tetration

[jovial rock]

Hmm

Ah

This cancels out

Nvm

[azure crane]

Btw $\ln_{\ln^s(a)}(a)=1$

[dawn finch]

RoyalBanana

[azure crane]

$\log_{a,{\log_a^s(b)}}(b)=1$

[dawn finch]

RoyalBanana

[azure crane]

$\ln_{\ln^s(a)}(b)={}^{ln^s(b)-ln^s(a)}e$

[dawn finch]

RoyalBanana

[jovial rock]

Not sure but I think x = ((e^^2)^^((x*(e^^(1-1/e))-e)/e)

[jovial rock]

Huh

I think I get it

Oh yeah that’s cool

[jovial rock]

Btw here we have at the start lns((e^^2)^^2

But I think we can make something like lns((e^^3)^^2

I think it won’t make a diffrence because lns(e^^x) cancels out

[jovial rock]

And for this substitution we could use 1 and e^^1-1/e —> e^^1-1 would cancel out to be 1

[jovial rock]

Wait

Is this even allowed with tetration?

On the left side

I’ll just take the take the lns

Nvm

Idk

Lemme find a general rule for (e^^a)^^b

[jovial rock]

Maybe with that

[jovial rock]

Ye I think the problem is that I used normal log

[jovial rock]

lns(ln(lns(a))(b)) = lns(b)-lns(a)

Hm

Hm, can we factor out the a from log(e^^a,-b)(1)

log(e^^a,-1)(1) is I think e^^a

[jovial rock]

rigth?

[jovial rock]

hmm

we could use that to climb down operations

[jovial rock]

lns(1) = lnx(1) = e^^-x
-lns(lns(1)) = x

[jovial rock]

ln(lns(a)-k)(a) = e^^k, but we need lns(lns(a)), because one lns cancels out and we get lns(a)

I think we would need some documentary where we can see all the rules

[azure crane]

NICE

[azure crane]

Do all of the steps again but replace every two with a three

nothing special about two there

[azure crane]

Yep

[azure crane]

ofc why not lol

[jovial rock]

ye but isn't it lns(e^^x) = x and not ln(e^^x) = x ?

[azure crane]

yes

?

why wouldn't that be allowed tho

$log_{({}^2e)}(\frac{1}{e})=-\frac{1}{e}$

[dawn finch]

RoyalBanana

[azure crane]

Just take e^^2 to the power of both sides

$\frac{1}{e}={(e^e)}^{-\frac{1}{e}}$

[dawn finch]

RoyalBanana

[jovial rock]

ye but I worry about the (e^^2)^^(...)

[dawn finch]

RoyalBanana

RoyalBanana

[azure crane]

Uh what?

[jovial rock]

(e^^2)^^(5) = 1/e

how to solve tahat

?

[azure crane]

solve for what

that's just a wrong equation lol

there's no variable to solve for lol

[azure crane]

Um I don't know what you mean

that is not super log it's normal log

[azure crane]

Nevermind I did it!

$x=({}^{\frac{x*{}^{(1-\frac{1}{{}^{(x-1)}e})}({}^xe)-e}{e}})({}^xe)$

[dawn finch]

RoyalBanana

[azure crane]

ahh latex

It can't put powers up that high no room

for all x not 1

Imma continue writing some lol

[jovial rock]

oh nicee

I was just really confused with the log of tetration

[azure crane]

$\log^s_{({}^ee)}(e)=\frac{1}{2}?$

[dawn finch]

RoyalBanana

[azure crane]

Maybe maybe not

[jovial rock]

Hmm

Maybe

[azure crane]

Ah no nvm

[azure crane]

${}^{({}^{(\frac{1}{2})}4)}4=16$

[dawn finch]

RoyalBanana

[azure crane]

I swapped e for 2

I don't see why not

[jovial rock]

Uhh

Lemme aprox that

I got 96.7532…

Not sure

Uhh

I have a question

What does log(e^^2)(e^^5) and what does log(e^^2)s(e^^5) equal?

[azure crane]

Idk the second one but I can do the first one

Oooh

I got a new formula

$\log_{({}^ax)}({}^bx)=\frac{{}^{(b-1)}x}{{}^{(a-1)}x}$

[dawn finch]

RoyalBanana

[azure crane]

$\log_{({}^2e)}({}^5e)=\frac{{}^{(5-1)}e}{{}^{(2-1)}e}$

[dawn finch]

RoyalBanana

[azure crane]

$=\frac{{}^{4}e}{e}$

[dawn finch]

RoyalBanana

[jovial rock]

this is true?

[azure crane]

yup

[jovial rock]

how do you know?

[azure crane]

100%

[jovial rock]

ah

uh

[azure crane]

Well other than the fact that every graph i checked for a bunch of as and bs works i think you can prove it with the loga(x)=ln(x)/ln(a)

[jovial rock]

oh okay

[azure crane]

Works with the switching log thing ofc also!

[jovial rock]

I mean this when the base is e^^2, from the ln and super ln

[jovial rock]

yeah true

[azure crane]

$\log_{({}^2e)}({}^5e)=\frac{{}^{4}e}{e}$

[dawn finch]

RoyalBanana

[azure crane]

idk about the superlog we'll figure it out lol

[jovial rock]

hm

[jovial rock]

I think the log(e^^2) I used here isn't right

[azure crane]

Wdym?

[jovial rock]

nvm

I don't understand what I did back then

[azure crane]

I couldn't find a mistake in it lol

[azure crane]

Don't forget the steps on the right

[jovial rock]

this one here

I don't think that log(e^^2) can cancel out the (e^^2)

because it's tetration

i think it would only work if it would have been eponetiation

[azure crane]

?

[jovial rock]

like (e^^2)^....

wouldn't you need log(e^^2)s

[azure crane]

By definition ${}^{(log^s_{{}^2e}(x))}({}^2e)=x$

[dawn finch]

RoyalBanana

[jovial rock]

I know, but this is normal log and not super log

[azure crane]

wheres normal log?

[jovial rock]

here

log(e^^2,1)()

[azure crane]

ohhhhhhhhhhh

:/

[jovial rock]

so we need to find out what the super log of 1/e is

and then we can solve this equation correctly

[azure crane]

Ok so instead of $-\frac{1}{e}$ it's $log^s_{{}^2e}(\frac{1}{e})$

[dawn finch]

RoyalBanana

[azure crane]

yeah

[jovial rock]

yep

right

[jovial rock]

thats why this doesn't work

[azure crane]

$2=({}^{\frac{2*{}^{(1+log^s_{{}^2e}(\frac{1}{e})}({}^2e)-e}{e}})({}^2e)$

[dawn finch]

RoyalBanana

[jovial rock]

log(x^^a,k)(x^^b) = (x^^(b-k))/(x^^(a-k))

[jovial rock]

yess

[azure crane]

?

[jovial rock]

it's just like your rule but just iterated so we can replace 1 with k

and we can get non integer values for k

[azure crane]

How's it iterated tho

It doesn't work for k=2

[jovial rock]

oh

wait

[azure crane]

Don't forget about the denominator

[jovial rock]

okay I see

you're right

[azure crane]

Although if a=1 then there's a rule there

Ah ofc $\log_{x,k}({}^bx)={}^{(b-k)}x$

[dawn finch]

RoyalBanana

[azure crane]

Btw should I make a word document for all the formulas

[jovial rock]

I already started a bit

but I don't have all of them now

[azure crane]

I can just look through the entire page

[jovial rock]

okay

[azure crane]

Btw you can use the ${}^{(log^s_a(x)-k)}a=log_{a,k}(x)$ formula here

[dawn finch]

RoyalBanana

[azure crane]

progress!

alr imma make the word document

[jovial rock]

okay, thanks

[azure crane]

scrolling through all of this is gonna be a pain but worth it lol

[jovial rock]

I'll try to understand how to combine these two equation

[jovial rock]

yeah true

from now on we can just add the equation after discovering it

[azure crane]

the e^^2^^(1+logs(e^^2)(1/e)) thing

Oh

I understand it now

interesting

[jovial rock]

hm

am confused

1+log(e^^2)(1/e) this is just 1 - 1/e, right?

[azure crane]

*logs

typo lol oops

[jovial rock]

huh

[jovial rock]

I thought this

[azure crane]

superlog not normal log

[jovial rock]

ahh

I see

super log

oke

[azure crane]

I fixed the typo

Imma make two versions of all the formulas one for base e and one for any base a bc e is special ofc with ln and stuff

Alright time to make the word document lol

[jovial rock]

okay, easyy

[azure crane]

This is probably the most useful one lol

[jovial rock]

yeah true

I like the one with 1/e too

[azure crane]

Man I wish this worked lol

[jovial rock]

because it allowes you to retransform the whole thing without breaking rules

yeahh, that one would have solved everything haha

[azure crane]

Alright I reached the top, finished downloading

Now to put them in a word document

About 19 formulas lol

Probably some duplicates there

[jovial rock]

yeah true

and not every is true

[jovial rock]

hm

[azure crane]

$\ln^s(a)=\ln_{(-\ln^s(\ln^s(a)))}(1)$

[dawn finch]

RoyalBanana

[jovial rock]

That works?

[azure crane]

yup

[jovial rock]

Ah right

Because ln(-k)(1) = e^^k

And if we want to have lns(a) = lnx(a) it would be ln(lns(a)-lns(lns(a)))(a)

[azure crane]

Yeah

I had that step

Wanna know the cool thing tho

Remember ln(a+b)(x)=lna(lnb(x))

$\ln_{(lns(a)-lns(lns(a)))}(a)=\ln_{(-lns(lns(a)))}(ln_{lns(a)}(a))$

[dawn finch]

RoyalBanana

[azure crane]

But ofc $\ln_{\ln^s(a)}(a)=1$

[dawn finch]

RoyalBanana

[azure crane]

So $\ln_{(lns(a)-lns(lns(a)))}(a)=\ln_{(-lns(lns(a)))}(ln_{lns(a)}(a))=\ln_{(-lns(lns(a)))}(1)$

[dawn finch]

RoyalBanana

[azure crane]

So $\ln^s(a)=\ln_{(-\ln^s(\ln^s(a)))}(1)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{(1+log^s_{{}^2e}(\frac{1}{e}))}({}^2e)=log_{({}^2e,-1)}(\frac{1}{e})$

[dawn finch]

RoyalBanana

[azure crane]

but ofc $log_{(a,-1)}(x)=a^x$

[dawn finch]

RoyalBanana

[azure crane]

So ${}^{(1+log^s_{{}^2e}(\frac{1}{e}))}({}^2e)=({}^2e)^{1/e}=(e^e)^{1/e}=e^{e*1/e}=e^1=e$

heck yes

[dawn finch]

RoyalBanana

[jovial rock]

Oh yeah, that’s nice. Can we solve it for lns(1) or log for every hyperoperation?

[azure crane]

not really

you can try i suppose

[jovial rock]

Oh wow

[azure crane]

:>

[azure crane]

Wait what? $2={}^{({}^{\frac{2*e-e}{e}})}({}^2e)$

[jovial rock]

So log(e^^2,-2)(1/e) = (e^^2)^^(1+log(e^^2)s(1/(e^e))

[dawn finch]

RoyalBanana

[azure crane]

$2={}^2e$

[dawn finch]

RoyalBanana

[jovial rock]

Doesn’t make sense

[azure crane]

uhh...

yeah idk about that

[jovial rock]

Wasn’t there the problem with log(e^^2)s(1/e) ?

[azure crane]

This is correct tho I just used the loga,k(x) formula

[azure crane]

This is correct

[azure crane]

So is this

[azure crane]

So this must be wrong :/

[jovial rock]

Hm

But what is wrong?

[azure crane]

I'm not sure

[jovial rock]

So ln(e^^2,-1)(1/e) is e, (e^^2,0)(1/e), so (e^^2,-2)(1/e) = e^2 ?

[azure crane]

do you mean log not ln lol

[jovial rock]

Oh ye sry haha

[azure crane]

you're good

$log_{({}^2e,0)}(\frac{1}{e})=\frac{1}{e}$

$log_{({}^2e,-1)}(\frac{1}{e})=e$

$log_{({}^2e,-2)}(\frac{1}{e})=?$

[dawn finch]

RoyalBanana

[jovial rock]

How to Write all of them in tetration form? (e^^2)^^…. I know -1 but not 0 or -2

[azure crane]

I assume it's e^e^^2=e^^3

Or is it (e^^2)^e?

$log_{({}^2e,-2)}(\frac{1}{e})=x$

$({}^2e)^{log_{({}^2e,-2)}(\frac{1}{e})}=({}^2e)^x$

$log_{({}^2e,-1)}(\frac{1}{e})=({}^2e)^x$

$e=({}^2e)^x$

[dawn finch]

RoyalBanana

[jovial rock]

I think it‘s log(e^^2,-k)(1/e) = (e^^2)^^(k+ log(e^^2)s(1/e))

[dawn finch]

RoyalBanana

[azure crane]

It's 1/e again! Weird

Wait does it just keep switching between e and 1/e?

[jovial rock]

I think not

[azure crane]

$log_{({}^2e,0)}(\frac{1}{e})=\frac{1}{e}$

$log_{({}^2e,-1)}(\frac{1}{e})=e$

$log_{({}^2e,-2)}(\frac{1}{e})=\frac{1}{e}$

[dawn finch]

RoyalBanana

[azure crane]

this is interesting lol

Ok so $log_{({}^2e,-3)}(\frac{1}{e})=log_{({}^2e,-1)}(log_{({}^2e,-2)}(\frac{1}{e}))$

[dawn finch]

RoyalBanana

[azure crane]

basic log rule

Ok so $log_{({}^2e,-3)}(\frac{1}{e})=log_{({}^2e,-1)}(1/e)$

[dawn finch]

RoyalBanana

[azure crane]

=e

wow

it really does go back and forth

interesting

[jovial rock]

log(e^^2,0)(1/e) = log(e^^2,-2)(1/e) i don‘t get it

[azure crane]

yup

weird

I know

I don't think any mistakes were made tho

Lemme double check on paper

[azure crane]

Ohhh I see the mistake canceling out the log and e^^2 would make the number of logs go down not up here

it would be -3

Not -1

ok I did it right this time it's e^^3

[jovial rock]

For -2 ?

[azure crane]

Yes

[jovial rock]

So it‘s 1/e, e, e^^3

How do you know it‘s e^^3 ?

[azure crane]

(e^^2)^the last one

[jovial rock]

(e^^2)^e ?

[azure crane]

yes

Basic log rules I just realized lol

[jovial rock]

(e^^2)^e ≠ e^(e^^2)

a^b ≠b^a

[azure crane]

(e^^2)^e=e^^3

Wait

No

[azure crane]

This is true tho

[azure crane]

*e^(e^2)

[jovial rock]

How

How did you got e^(e^^2) ?

[azure crane]

I didn't I got e^(e^2)

[jovial rock]

Ah

Okay

[azure crane]

$({}^2e)^e=(e^e)^e=e^{(e*e)}=e^{(e^2)}$

[dawn finch]

RoyalBanana

[jovial rock]

Okay, I See

Isn‘t this log(e^^2,-2)(1/e) = (e^^2)^^(2+log(e^^2)s(1/(e))

[azure crane]

?

[jovial rock]

log(e^^2,-1)(1/e) = (e^^2)^^(log(e^^2)s((e^^2)^(1/(e))) right?

[azure crane]

Yes

btw (e^^2)^(1/e)=e

[jovial rock]

So log(e^^2,-2)(1/e) = (e^^2)^^(log(e^^2)s((e^^2)^(e^^2)^(1/(e))) right?

Right?

[jovial rock]

Yee. True

[azure crane]

Yes

[jovial rock]

question. a^^^k = ln5(-k)(1)

[azure crane]

oh solve for a?

[jovial rock]

And then you Can factor out the e^^2 to get: log(e^^2,-2)(1/e) = (e^^2)^^(2+log(e^^2)s(1/(e))

[jovial rock]

ln5 is pentation ln

Oh wait

It‘s 4

ln4

[azure crane]

you gotta say ln^5 otherwise it's confusing

[azure crane]

nono that's tetration

[jovial rock]

ik but we used ln^3 for tetration

[azure crane]

?

no

[jovial rock]

e^^k = ln^3(-k)(1)

Thats by Definition of our formula

[azure crane]

^

It was based on the H function

4 is tetration

3 is exponentiation

2 is multiplication

1 is addition

[jovial rock]

ik, but I mean we used ln^3 to calculate tetration

ln^3 is normal ln

[azure crane]

ln^4 is tetration ln^3 is just ln

[jovial rock]

Yes

[azure crane]

Wdym calculate tetration

[jovial rock]

This

[azure crane]

yes

Ah

[jovial rock]

Is this rule for every operation?

[azure crane]

Yes

By definition

[jovial rock]

Operation -1 haha

ln^-1

[azure crane]

lnk is the inverse of exponentiation and ln-k is the inverse of lnk so ln-k is exponentiation

[jovial rock]

Yeah, true

[dawn finch]

RoyalBanana

[jovial rock]

Yess

a*k = log^1(-k)(1)

a+k = log^0(-k)(1)

[azure crane]

uhh

[jovial rock]

Over complicating this stuff but using the same rule

I wanna see log^-1(-k)(1)

[azure crane]

that's just log^0(-k)(1)

[jovial rock]

how do you know

[azure crane]

Remembering

[jovial rock]

The n = Rule works only for integer

[azure crane]

H(-1,a,b)=H(0,a,b)=b+1

remember

[azure crane]

just bc it's weird doesn't mean it's wrong

[jovial rock]

yeah but if you would repeat H(-1,a,b) you would get addition and this doesn‘t make sense for H(0,a,b)

[azure crane]

how does it not?

[jovial rock]

H(0,a,b) is succesion by defenition

[azure crane]

yes

[jovial rock]

And succesion ≠ addition

[azure crane]

no

That formula is the succession rule

[jovial rock]

Ye

Soo

So when n<=0 it‘s succession right?

If yes, -0.1 would be succession too, this repeated is 0.9

[azure crane]

Remember this H(n,a,b)=H(n-1,a,H(n-1,a,...H(n-1,a,a)))
b-1 Hs being applied on the right side

does this make sense

H(n,a,4)=H(n-1,a,H(n-1,a,H(n-1,a,a)))

[jovial rock]

I just don‘t get this here: H(-1,a,b) = succsession, and H(0,a,b) = succesion, and H(-1,a,b) repeated is addition, because repeated succesion is addition, and this would be addition = H(0,a,b) which is succession by definition, and this means addition = succesion and this is non sense

[azure crane]

What is succession

[jovial rock]

b+1

Adding

[azure crane]

That is addition by 1

[jovial rock]

Counting

[jovial rock]

Yes

[azure crane]

that is still addition tho

Just by 1

[jovial rock]

And ignoring a

[azure crane]

examples
n=2: e×4=e+(e+(e+e))
n=3: e^4=e×(e×(e×e))
n=4: e^^4=e^(e^(e^e))
n=5 e^^^4=e^^(e^^(e^^e))
Yes?

[jovial rock]

Yes makes sense

[azure crane]

now plug in n=1 ye?

e+4=e+(1+(1+1))

[jovial rock]

Ye

[azure crane]

Now plug in n=0

[jovial rock]

Hm

[azure crane]

H(0,a,2)=H(-1,a,a)

[jovial rock]

Uh

What would H(-1,a,2) be?

[azure crane]

idk

[jovial rock]

Yess

I like this more

[azure crane]

figure this out first

what is H(0,a,2)

[jovial rock]

a+1+1

S(S(a))

[azure crane]

so a+2=H(-1,a,a)

alr

you have convinced me lol

[jovial rock]

Yaaaaaa

[azure crane]

Lemme write the examples so this makes sense

[jovial rock]

I was just very unsatisfied with H(0,a,b) = H(-1,a,b) and i like this new example more

Because it gives more unsolved Problems

[azure crane]

H(n,a,2)=H(n-1,a,a)
n=5: a^^^2=a^^a
n=4: a^^2=a^a
n=3: a^2=a×a
n=2: a×2=a+a
n=1: a+2=a?a

some symbol

[jovial rock]

Ye

[azure crane]

n=0: a?2=uhh..

[jovial rock]

lol

[azure crane]

Imma look into that symbol some more

H(n,a,3)=H(n-1,a,H(n-1,a,a))

n=1: a+3=a?(a?a)

So a+3=a?(a+2)

interesting

H(n,a,4)=H(n-1,a,H(n-1,a,H(n-1,a,a)))

n=1: a+4=a?(a?(a?a))

so a+4=a?(a?(a+2))

so a+4=a?(a+3)

[jovial rock]

This operation would allow us to take to whole mathematical objects together and add them I guess..

[azure crane]

clearly a+b=a?(a+b-1)

let's replace b with b-a+1

a+b-a+1=a?(a+b-a+1-1)

b+1=a?b

there, proof

alr

now let's do the same for n=0

imma use ¿

ig

H(n,a,2)=H(n-1,a,a)

n=0: a?2=a¿a

3=a¿a

H(n,a,3)=H(n-1,a,H(n-1,a,a))

n=0: a?3=a¿(a¿a)

4=a¿(3)

H(n,a,4)=H(n-1,a,H(n-1,a,H(n-1,a,a)))

Wait

wait

what

uhh

[azure crane]

contradiction

that does not match with a+2=a?a

I am confusion

that's saying that a+2=a+1

uh..

Imma redo this on paper

[azure crane]

Would you say this is definitely correct though? It's kinda the definition of moving to the next higher/lower operation right?

@jovial rock

alr

[azure crane]

OHHHH I see where I went wrong

See normally like
a*1=a
a^1=a
a^^1=a
a^^^1=a

but $a+1 \ne a$

[dawn finch]

RoyalBanana

[azure crane]

It's the only one

with this property

which messes up a?b

a?b is like discontinuous somehow

it's not exactly b+1

$a?a \ne a+1, a?a=a+2$

[dawn finch]

RoyalBanana

[azure crane]

Wait

I have a theory...

this is weird

[jovial rock]

I think Yes

Sorry had to do some stuff

[azure crane]

Wait what's lns(0)?

[azure crane]

-infinity?

[jovial rock]

A number that isn‘t defined I guess

Yes

[azure crane]

Hehe my theory is working

[jovial rock]

I have a theory too

Not sure if this is a theory

[azure crane]

There is a more generalized version of this i just realized while making a table of numbers

this is weird

one second

idk how to word this

[jovial rock]

Isn‘t a?a = a+1?

[jovial rock]

Oh, Sounds intressting

[azure crane]

by definition no

[jovial rock]

Okay

[azure crane]

Remember ^

[jovial rock]

Ye

[azure crane]

It's weird

[azure crane]

^

[jovial rock]

Hm

Could be something

[azure crane]

It's almost b+1

not for some values

[jovial rock]

My theory was that a?1 is not a number as we know, it‘s more Like a mumber in a more abstact number class than the neutral numbers. And as soon you do a?a neutral Numbers get created

It‘s weird ik

[azure crane]

that's more philosophical than mathematical lol

[jovial rock]

Ye

[azure crane]

I kinda get it

[jovial rock]

I guess

[azure crane]

btw I got like 2?5 from the fact that
a+3=a?(a?a)=a?(a+2)

[jovial rock]

Maybe it could work similar to 1x2 = 1x3

[azure crane]

wait 2=3?

[jovial rock]

Oh

Sry

[azure crane]

you're good!

anyway I need an inverse for ?

[jovial rock]

I meant 1^2 =1^3

[azure crane]

like theres - for + ,÷ for × ,log and sqrt for ^

[jovial rock]

log^-1

Haha

lol

[azure crane]

lol

[jovial rock]

Idk

Hmm

[azure crane]

Imma actually steal the ¿ symbol for this

[jovial rock]

Okay, nice

[azure crane]

Let's see the usual definition for inverses

[jovial rock]

I mean there are going to be infinite symbols when it‘s operation -476

I guess a?x = b

But there is x?a = b too

To solve for x

But a?x = b is more useful

Because we use logarithm more often in this kind of math

[azure crane]

b¿(a?b)=a

[jovial rock]

Hmm

Lemme activate my Brain

[azure crane]

a¿(a+2)=a

I switched it

[jovial rock]

Yeah

Thats better

That was the Same thing with tetration log

I still feel a Bit insecure about a?a = a+2 but I guess I try to accept it because you Said it‘s by definition

[azure crane]

That's by the definition of going to the lower arithmetic operation

[jovial rock]

But at the start you said H(-1,a,b) = H(0,a,b)

[azure crane]

That was wrong

Like you said

But!

[jovial rock]

Okay

Lets accept a?a

[azure crane]

a^^^^2=a^^^a
a^^^2=a^^a
a^^2=a^a
a^2=a×a
a×2=a+a
a+2=a?a
does this explain it

a+3=a?(a?a)

[jovial rock]

Yeah it does, but when we Cross the addition line, iteration complexity gets inversed

[azure crane]

=a?(a+2)

[jovial rock]

Ye

[azure crane]

Ofc it's weird lol ye :>

[jovial rock]

I‘ll trust you with a?a = a+2 for now

[azure crane]

I mean

does the pattern not make sense lol

[jovial rock]

It does but still, it‘s Kinds weird

What does the operator ? Even mean

We‘re kinda going beyond the borders mathematics should supossed to have

[azure crane]

it's just a new symbol like × or + or ^

[jovial rock]

Yeah I know, but what it does

[azure crane]

stuff

[jovial rock]

Yes

But this stuff is really cool

A general formula for everything would be crazy, ngl

Like for the arithmetic stuff

[azure crane]

lol

[jovial rock]

Every algebraic equation could easily be solved by just using some crazy inverse

[azure crane]

..I mean you could argue the same thing that the formula only holds for integers or whatever

but it's really subjective what formula you use

[jovial rock]

Yeah, true

I think we should just continue with the things we think and then we‘ll find it out

[azure crane]

1?2=?

[jovial rock]

Hm

Wasn‘t there this one rule

Wait

[azure crane]

Imma repost these so you don't have to scroll up

H(n,a,b)=H(n-1,a,H(n-1,a,...H(n-1,a,a)))
b-1 Hs being applied on the right side

[jovial rock]

ln^4(a) = ln^3(-ln^4(ln^4(a)))(1)

we can climb down operations with this

I think

[azure crane]

I guess

But what is ln^2(a)

÷?

[jovial rock]

ye

[azure crane]

but by what

it depends on the base

[jovial rock]

Wait

e/x = a

Right?

[azure crane]

ohh $log^2_b(a)=\frac{a}{b}?$

[dawn finch]

RoyalBanana

[azure crane]

$log^1_b(a)=a-b$

[dawn finch]

RoyalBanana

[azure crane]

log^0_b(a)=a¿b

[jovial rock]

log^1(a) = log^0(-log^1(log^1(a)))(1)

[azure crane]

$log^n_b(H(n,a,b))=a$

[dawn finch]

RoyalBanana

[azure crane]

by definition

Wait

no

$log^n_a(H(n,a,b))=b$

[dawn finch]

RoyalBanana

[azure crane]

Yes

[jovial rock]

a-b = a¿(-(a-(a-b)))(1)

Right?

It‘s just a iteration thingy

[azure crane]

Can you show this works for stuff other than n=4

try 3 first

[jovial rock]

Wait

[azure crane]

$ln^4(a) = ln^3_{(-ln^4(ln^4(a)))}(1)$

[dawn finch]

RoyalBanana

[azure crane]

$ln(a) = ln^2_{(-ln(ln(a)))}(1)$?

[dawn finch]

RoyalBanana

[azure crane]

Lemme use this to figure out ln^2

$log_a^2(ab))=b$

[dawn finch]

RoyalBanana

[azure crane]

Ah I seee

$log_y^2(x)=\frac{x}{y}$

[jovial rock]

Ye

[dawn finch]

RoyalBanana

[azure crane]

I thought so lol

Alr so

$ln(a) = ln^2_{(-ln(ln(a)))}(1)$?

[dawn finch]

RoyalBanana

[jovial rock]

Uh

The other way

[azure crane]

Ah but what's $ln^2_k(x)$?

[dawn finch]

RoyalBanana

[jovial rock]

ln^2(a) = ln(….

[azure crane]

That would give ln^1

[jovial rock]

x-k

[azure crane]

besides this should work too

[jovial rock]

Ye

[azure crane]

Wrong one, if k=1 it doesn't work

[jovial rock]

Uh

Ah

It‘s ln

Aha

And it uses k

x-e*k

Right?

[azure crane]

Remember $ln^n_a(ln^n_b(x))=ln^n_{(a+b)}(x)$

$ln^2_1(x)=x/e$

$ln^2_2(x)=x/e^2$

$ln^2_3(x)=x/e^3$

[dawn finch]

RoyalBanana

[jovial rock]

Aha

[azure crane]

it just keeps dividing by e each time

[jovial rock]

Yeah

[azure crane]

$ln^2_k(x)=x/e^k$

[dawn finch]

RoyalBanana

[azure crane]

extension to non integers ofc!

[jovial rock]

Btw I think we can‘t do this because the non integer k

[jovial rock]

Yeah

[azure crane]

Okok so

$ln(a) = ln^2_{(-ln(ln(a)))}(1)$?

[dawn finch]

RoyalBanana

[azure crane]

let's do that after ig

[jovial rock]

Okay

[jovial rock]

Hmm

[azure crane]

$ln(a) = ln^2_{(-ln(ln(a)))}(1)=\frac{1}{e^{-ln(ln(a))}}?$

[dawn finch]

RoyalBanana

[jovial rock]

You mean ln^1 right?