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

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

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

We move all terms to the left:

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


Domain of the equation: 4^(2-3x)+2(3x+3))!=0

x∈R


We add all the numbers together, and all the variables

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

We calculate fractions


2^(3x+1)+()/(4^(-3x+2)+2(3x+3))*4^()+(4x(-6x+)/(4^(-3x+2)+2(3x+3))*4^()=0


We calculate terms in parentheses: +()/(4^(-3x+2)+2(3x+3))*4^(), so:

)/(4^(-3x+2)+2(3x+3))*4^(

We multiply all the terms by the denominator


)

We add all the numbers together, and all the variables



Back to the equation:

+()



We calculate terms in parentheses: +(4x(-6x+)/(4^(-3x+2)+2(3x+3))*4^(), so:

4x(-6x+)/(4^(-3x+2)+2(3x+3))*4^(

We add all the numbers together, and all the variables

4x(-6x)/(4^(-3x+2)+2(3x+3))*4^(

We multiply all the terms by the denominator


4x(-6x)

We multiply parentheses

-24x^2

Back to the equation:

+(-24x^2)


We add all the numbers together, and all the variables

(-24x^2)+2^(3x+1)=0

We get rid of parentheses

-24x^2+2^(3x+1)=0