22=(1/5)^2x+7

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Solution for 22=(1/5)^2x+7 equation: 



22=(1/5)^2x+7

We move all terms to the left:

22-((1/5)^2x+7)=0



Domain of the equation: 5)^2x+7)!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

-((+1/5)^2x+7)+22=0

We multiply all the terms by the denominator


-((+1+22*5)^2x+7)=0



We calculate terms in parentheses: -((+1+22*5)^2x+7), so:

(+1+22*5)^2x+7

We add all the numbers together, and all the variables

111^2x+7

Back to the equation:

-(111^2x+7)



We get rid of parentheses

-111^2x-7=0

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

-111^2x=7

x=7/1

x=7

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