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

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



=(2/3)^Y+4-2

We move all terms to the left:

-((2/3)^Y+4-2)=0



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

Y!=0/1

Y!=0

Y∈R



We add all the numbers together, and all the variables

-((+2/3)^Y+4-2)=0

We multiply all the terms by the denominator


-((+2=0

We add all the numbers together, and all the variables

=0

Y=0/1

Y=0

Równanie nie ma rozwiązania

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