3^4-5=(1/27)^2x+10

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Solution for 3^4-5=(1/27)^2x+10 equation: 



3^4-5=(1/27)^2x+10

We move all terms to the left:

3^4-5-((1/27)^2x+10)=0



Domain of the equation: 27)^2x+10)!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

-((+1/27)^2x+10)-5+3^4=0

We add all the numbers together, and all the variables

-((+1/27)^2x+10)+76=0

We multiply all the terms by the denominator


-((+1+76*27)^2x+10)=0



We calculate terms in parentheses: -((+1+76*27)^2x+10), so:

(+1+76*27)^2x+10

We add all the numbers together, and all the variables

2053^2x+10

Back to the equation:

-(2053^2x+10)



We get rid of parentheses

-2053^2x-10=0

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

-2053^2x=10

x=10/1

x=10

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