Wait, is this true?

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You are currently reading a thread in /sci/ - Science & Math

You are currently reading a thread in /sci/ - Science & Math

Thread images: 2

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.

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>>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]

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>>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]

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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.

>>

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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>>7809075

Show me how

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>>7809018

What was hard about that?

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.

>>

>>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.

>>

>>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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