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Brainlet Question

Anonymous Thu 28 Nov 11:00:55 2019 No.11183033 View Reply Original Report

Quoted By: >>11183048 >>11183053 >>11184020 >>11184084 >>11185430

Brainlet Question: If roots are the inverse operations of exponents, why are the inverse functions of exponential functions logarithms and not root?

Anonymous

Anonymous Thu 28 Nov 2019 11:16:20 No.11183048 Report

Quoted By: >>11184137

>>11183033
There's a difference between x^a and a^x.

Anonymous

Anonymous Thu 28 Nov 2019 11:21:40 No.11183053 Report

Quoted By: >>11183336

>>11183033
x^n is not exponential, n^x is. it's inverse is logarithmic function.

x^n is just polinomial, it's inverse is the root.

Anonymous

Anonymous Thu 28 Nov 2019 13:43:22 No.11183320 Report

Quoted By:

Shut up Jew

Anonymous

Anonymous Thu 28 Nov 2019 13:50:06 No.11183336 Report

Quoted By: >>11183539

>>11183053
x^n has n values of x leading to the desired result. The corresponding root function only yields 1.

Justin Ramsey

Justin Ramsey Thu 28 Nov 2019 14:33:08 No.11183376 Report

Quoted By:

Clasp Infinite 0s Does Not Process

Justin Ramsey

Justin Ramsey Thu 28 Nov 2019 14:35:01 No.11183380 Report

Quoted By:

The Best Scare Tactics To The Brain Damaged Will Appear On The Internet Cryptically

Anonymous

Anonymous Thu 28 Nov 2019 16:06:37 No.11183539 Report

Quoted By: >>11183609 >>11183780

>>11183336
The root function can only produce 1 by the very definition of a function. You can get all of them by allowing it to be a multifunction.

Anonymous

Anonymous Thu 28 Nov 2019 17:00:04 No.11183609 Report

Quoted By:

>>11183539
Or by solving the polynomial.

Anonymous

Anonymous Thu 28 Nov 2019 18:29:15 No.11183780 Report

Quoted By: >>11183958

>>11183539
Wtf is a multifunction? A function is by definition a well-defined relation. It can not have 2 values.

Anonymous

Anonymous Thu 28 Nov 2019 19:37:12 No.11183958 Report

Quoted By:

>>11183780
A multifunction (or multi-valued function) is not a function. It's a left-total relation, given the name of "function" because of its use in extension in complex analysis.

Anonymous

Anonymous Thu 28 Nov 2019 20:01:37 No.11184020 Report

Quoted By:

>>11183033
I think you got lost baffling yourselves with bullshit.

Anonymous

Anonymous Thu 28 Nov 2019 20:28:51 No.11184084 Report

Quoted By: >>11184156

>>11183033
Exponentiating very often is not injective (therefore not bijective), so you don't have proper inversion functions

Anonymous

Anonymous Thu 28 Nov 2019 20:46:31 No.11184137 Report

Quoted By: >>11184282 >>11185994

>>11183048
umm wrong, 2^2 = 2^2

get absolutely btfo.

Anonymous

Anonymous Thu 28 Nov 2019 20:53:37 No.11184156 Report

Quoted By:

>>11184084
If a is a positive real number which is not one, a^x is injective.

Anonymous

Anonymous Thu 28 Nov 2019 22:01:30 No.11184282 Report

Quoted By:

>>11184137
They're not the same, Anon. Don't you see? Look closer.

Anonymous

Anonymous Fri 29 Nov 2019 09:02:52 No.11185430 Report

Quoted By:

>>11183033
find the inverse of x^x as a nice little exercise.
also just as a bonus, which should be trivial, find x, such that x^x = 2

Anonymous

Anonymous Fri 29 Nov 2019 16:55:33 No.11185994 Report

Quoted By:

>>11184137
Shit tier, next time use 2^4 = 4^2

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