(1/9)^5x+1/2=27^(3x-2

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Solution for (1/9)^5x+1/2=27^(3x-2 equation: 



(1/9)^5x+1/2=27^(3x-2

We move all terms to the left:

(1/9)^5x+1/2-(27^(3x-2)=0



Domain of the equation: 9)^5x!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

(+1/9)^5x-(27^(3x-2)+1/2=0

We calculate fractions


()/9^5x*2)+(-(27^(3x-2)+1*9)^5x)/9^5x*2)=0



We calculate terms in parentheses: -(27^(3x-2)+1*9)^5x)/9^5x*2), so:

27^(3x-2)+1*9)^5x)/9^5x*2

We multiply all the terms by the denominator


(27^(3x-2))*9^5x*2+1*9)^5x)

Wy multiply elements

(27^(3x-2))*9^5x*2+9x

We add all the numbers together, and all the variables

9x+(27^(3x-2))*9^5x*2

Back to the equation:

-(9x+(27^(3x-2))*9^5x*2)



We multiply all the terms by the denominator


-((9x+(27^(3x-2))*9^5x*2))*9^5x*2)+(+()=0

We add all the numbers together, and all the variables

-((9x+(27^(3x-2))*9^5x*2))*9^5x*2)+(=0

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