№4083130[Quote]
y = mx+b
№4083134[Quote]
okay, let's start with x = 1 and y = 1
№4083145[Quote]
>>4083134This won't suffice
№4083148[Quote]
x^(x + y) = y
№4083149[Quote]
>>4083134X and Y are different numbers you stupid nigger
№4083154[Quote]
>>4083146Correct.
There's also
x = -1
y = 1
№4083165[Quote]
tricky, huh
№4083167[Quote]
>>4083124 (OP)Infinite solutions.
№4083188[Quote]
>>4083174What values for x and y do you allow?
№4083191[Quote]
File: NWO.PNG 📥︎ (212.94 KB, 1593x1664) 07047969462c48b394769c18a1eefaac0ImgOps
>>4083154>There's also>x = -1>y = 1A logarithm can't have a negative base unless you expand the domain to allow for complex numbers though.
№4083192[Quote]
>>4083188any that suffice x + y = log_x(y) or x^(x + y) = y
№4083194[Quote]
>>4083154Invalid log_(-1) is not a well defined function.
№4083197[Quote]
>>4083191We can work with complex numbers
№4083199[Quote]
>>4083192Do you allow y to be complex that is what I am asking.
№4083208[Quote]
>>4083201Okay, otherwise I do not know if there are infinite solutions, but log_x is a periodic function on C so if you have one solution you can rotate by 2*pi*i
№4083231[Quote]
x+y = log_x(y)
Setting x = e we can get:
(e^y)/y= e^e.
This definitely will have a solution I am sure. But it is unlikely that you will be able to calculate the exact value.
№4083245[Quote]
>>4083240Do you even know what log means?
№4083247[Quote]
Btw I checked with a graph calulator. It showed to solutions. One around 4. something, I am very sure there is no closed form.
№4083250[Quote]
>>4083245It's an incognito exponent right? log (Ie: Exponent) of X base which equals Y, right??
№4083252[Quote]
Wait I got it.
№4083254[Quote]
>>4083124 (OP)Nigger at least give X or y.
How any one spouse to get it? this lie asking if x+y=xy*3
№4083261[Quote]
>>4083250No log solves the equation:
e^x = 4
for example, so log_x(y) is to say
x^(x+y) =y
№4083264[Quote]
>>4083252You can solve via the lambert W function.
№4083267[Quote]
A solution where both numbers are complex / imaginary
x = i
y = z - i
Where i^z = z - i
№4083274[Quote]
>>4083267Which is not solveable I think
№4083275[Quote]
>>4083267This is not true.
№4083282[Quote]
log to base i is not defined. x has to be real no matter what, y can be anything.
№4083296[Quote]
>>4083261Ohhhhh, only log equals x+y? I think I misunderstood because of text formating.
№4083298[Quote]
Is that what the underline is for?
№4083304[Quote]
>>4083286The only chance I see is probably some lambert W trickery, but again that would not be a closed solution. Btw x=2.71…=e and y =4.1… solves your question.
№4083305[Quote]
>>4083296Think of log_x(y) as a function that returns such number z, that x^z = y, defined only across real numbers apparently.
№4083320[Quote]
>>4083305The reason x can only be a positive real number is because you want it to be a continous function.
№4083321[Quote]
>>4083318yes, in this case.
№4083322[Quote]
>>4083321HOLY SHIT I UNDERSTAND IT
№4083330[Quote]
>>4083320It's reasonable.