49^x-5=(1/7)^4x

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Solution for 49^x-5=(1/7)^4x equation: 



49^x-5=(1/7)^4x

We move all terms to the left:

49^x-5-((1/7)^4x)=0



Domain of the equation: 7)^4x)!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

49^x-((+1/7)^4x)-5=0

We multiply all the terms by the denominator


49^x*7)^4x)-((-5*7)^4x)+1=0

We add all the numbers together, and all the variables

49^x*7)^4x)-((-35)^4x)+1=0

Wy multiply elements

343x^2-35)^4x)+1=0

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