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Wait, is this true?
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You are currently reading a thread in /sci/ - Science & Math

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

$2^x + 3^x = 11$ without it. You need to use Taylor Series yes?

This equation kind of circumvents that, to an extent.
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>>7809002
Yes.
Because
$\frac{log(1 + x^{ \frac {m log(y) } { logx } - n })}{log(x)} = log_x(1+log_x (y) - n)$
so
$x^{n+log_x(1+x^{ m log_x(y) -n}= x^n*(1+\frac{y^m}{x^n}$
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>>7809021
fixed last line:
$x^{n + log_{x} ( 1 + x ^ { m log_{x}(y) - n})}= x ^ n*(1 + \frac{ y^m }{ x^n})$
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>>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.
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This formula looks shitty and useless. Prove me wrong
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>>7809002
gr8 b8 m8
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>>7809018
isn't it in that video about fucking "Is it racist?"
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>>7809060
You can use it here: >>7809018
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>>7809075
Show me how
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>>7809018

http://www.wolframalpha.com/input/?i=2%5Ex%2B3%5Ex%3D11
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>>7809018
Try to solve it with it
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>>7809002
My intuition dictates that the complex parenthetical there can be given a simpler notation.
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>>7809092
>Show me how
>me unable to plug-n-chug
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>>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 ...
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>>7809476
>complex parenthetical
>real numbers
fgt pls
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>>7809002
I derived this before and yes you use two change of bases for the logarithms to work thi sout.
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>>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.
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>>7810928
i asked because its impossible to do it and want to see where >>7809018 made the error that led him to think it works