fuckong math

>homework
Go to /wsr newfag

wat is a log

>>8491716
>x + 3^x < 4
lol naplet.

x+3^x = (1 + 3^)x
-> x < (1+3^)^-1 * 4 =(1 + 1/3) * 4 = 5 + 1/3

on a serious note:
x + 3^x = 4 for x = 1. the function is strictly monoton therefore for any x < 1 it is < 4

you can use Lambert W to solve exactly

>>8491742
This is a log

>>8491750
what is
>3^)x

>>8491760
notation for the outer expontial as defined in the complex differential geometrics which is needed for advanced teichmuller theory.

>>8491716

x + x^3 = 4 => /derivate/
1 + 3x^2 = 0 =>
X^2 = -1/3 => /insert in original/
2x/3 = 4 => x = 6

File: tfw you study complex analysis for three years and finally understand what that means.png (3KB, 147x262px) Image search: [Google]

3KB, 147x262px

>>8491769
(You)

>>8491769
I know some basics but still don't understand how did u use binary operation on R without specifying the second argument of operation.

the proof is obviously trivial if x is at most a rational number, but is it actually possible to do it using P1 - P12 in the case of the real numbers?
I just dont see how its possible without having
a^x = e^xlog a defined.

>>8491790
It says 3^x, not the other way around.

i think the purpose of the question is to make the student waste time and realize that the problem if unsolvable with precalc methods

0 kek

the function [math] x+3^x[/math] is monotonously increasing, so you only have to find the point where [math] x+3^x=4[/math]
by inspection, that is 1. you could have approximated this numerical methods, dunno if there is a general method to get such a solution tho

Pol here. If x= 1 than the result of your equation is 4. 4 is not < 4 . So x is < 1

Did I dun good?