(4/3-4)n=(2/3+2)n

Solution for (4/3-4)n=(2/3+2)n equation:

(4/3-4)n=(2/3+2)n

We move all terms to the left:

(4/3-4)n-((2/3+2)n)=0


Domain of the equation: 3-4)n!=0

We move all terms containing n to the left, all other terms to the right

-4)n!=-3

n!=-3/1

n!=-3

n∈R



Domain of the equation: 3+2)n)!=0

We move all terms containing n to the left, all other terms to the right

2)n)!=-3

n!=-3/1

n!=-3

n∈R


We multiply parentheses

4n^2-4n-((2/3+2)n)=0

We multiply all the terms by the denominator


4n^2*3+2)n)-4n*3+2)n)-((2=0

Wy multiply elements

12n^2+2)n)-4n*3+2)n)-((2=0