Wait, is this true?
It is supposedly true for all x, y, n, and m that are real.
>>7809002
This is actually kind of a big deal guys. Try to solve the equation
[math] 2^x + 3^x = 11[/math] without it. You need to use Taylor Series yes?
This equation kind of circumvents that, to an extent.
Some youtube guy from London claims to have made it.
>>7809002
Yes.
Because
[math]
\frac{log(1 + x^{ \frac {m log(y) } { logx } - n })}{log(x)} = log_x(1+log_x (y) - n)
[/math]
so
[math]
x^{n+log_x(1+x^{ m log_x(y) -n}= x^n*(1+\frac{y^m}{x^n}
[/math]
>>7809021
fixed last line:
[math]
x^{n + log_{x} ( 1 + x ^ { m log_{x}(y) - n})}= x ^ n*(1 + \frac{ y^m }{ x^n})
[/math]
>>7809018
This is the only kind of elementary algebra equation (no differentials) that I thought required calculus to solve and isn't taught in high schools.
I didn't know there was a formula for it. Same way they don't teach high schoolers about the quartic or triatic formulas I guess, instead just waiting for them to learn about Galois.
This formula looks shitty and useless. Prove me wrong
>>7809002
gr8 b8 m8
>>7809018
isn't it in that video about fucking "Is it racist?"
>>7809075
Show me how
>>7809018
What was hard about that?
http://www.wolframalpha.com/input/?i=2%5Ex%2B3%5Ex%3D11
>>7809018
Try to solve it with it
>>7809002
My intuition dictates that the complex parenthetical there can be given a simpler notation.
>>7809092
>Show me how
>me unable to plug-n-chug
>>7809002
well it does not work for X<=0 or Y<=0.
Apart from that, it seems to be working ... but it does not look helpful to solve the 2^n+3^n=11 type of equations ...
>>7809476
>complex parenthetical
>real numbers
fgt pls
>>7809002
I derived this before and yes you use two change of bases for the logarithms to work thi sout.
>>7811144
>2^n+3^n=11 type of equations ...
thats because it only solves x^n + y^m = x^a
where you solve for a.