(1/5)^2x+4=125^x

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Solution for (1/5)^2x+4=125^x equation: 



(1/5)^2x+4=125^x

We move all terms to the left:

(1/5)^2x+4-(125^x)=0



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

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

(+1/5)^2x-125^x+4=0

We multiply all the terms by the denominator


(+1-125^x*5)^2x+4*5)^2x=0

Wy multiply elements

(+1-125^x*5)^2x+20x=0

We add all the numbers together, and all the variables

20x+(+1-125^x*5)^2x=0

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