(1/3)^5x+6=9x-1

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Solution for (1/3)^5x+6=9x-1 equation: 



(1/3)^5x+6=9x-1

We move all terms to the left:

(1/3)^5x+6-(9x-1)=0



Domain of the equation: 3)^5x!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

(+1/3)^5x-(9x-1)+6=0

We get rid of parentheses

(+1/3)^5x-9x+1+6=0

We multiply all the terms by the denominator


(+1-9x*3)^5x+1*3)^5x+6*3)^5x=0

We add all the numbers together, and all the variables

(-9x*3+1)^5x+1*3)^5x+6*3)^5x=0

Wy multiply elements

9x^2+(-9x*3+1)^5x=0

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