(80*0.40)^3/(40*1.6)^2*(128)^4=(2)^x

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Solution for (80*0.40)^3/(40*1.6)^2*(128)^4=(2)^x equation: 



(80*0.40)^3/(40*1.6)^2(128)^4=(2)^x

We move all terms to the left:

(80*0.40)^3/(40*1.6)^2(128)^4-((2)^x)=0

determiningTheFunctionDomain
-2^x+(80*0.40)^3/(40*1.6)^2128^4=0

We add all the numbers together, and all the variables

-2^x+32^3/64^2128^4=0

We multiply all the terms by the denominator


-2^x*64^2128^4+32^3=0

We add all the numbers together, and all the variables

-2^x*64^2128^4+32768=0

Wy multiply elements

-128x^2129+32768=0

We do not support expression: x^2129

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