3^4x-5=(1/7)^2x+10

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Solution for 3^4x-5=(1/7)^2x+10 equation: 



3^4x-5=(1/7)^2x+10

We move all terms to the left:

3^4x-5-((1/7)^2x+10)=0



Domain of the equation: 7)^2x+10)!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

3^4x-((+1/7)^2x+10)-5=0

We multiply all the terms by the denominator


3^4x*7)^2x+10)-((-5*7)^2x+10)+1=0

We add all the numbers together, and all the variables

3^4x*7)^2x+10)-((-35)^2x+10)+1=0

Wy multiply elements

21x^5-35)^2x+10)+1=0

We do not support expression: x^5

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