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1(1/2)^x=1/(2^x)
We move all terms to the left:
1(1/2)^x-(1/(2^x))=0
Domain of the equation: 2)^x!=0
x!=0/1
x!=0
x∈R
Domain of the equation: 2^x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1(+1/2)^x-(1/2^x)=0
We get rid of parentheses
1(+1/2)^x-1/2^x=0
We calculate fractions
(1(+1*2^x)/2^x*2^x)+(-2x)/2^x*2^x)=0
We calculate terms in parentheses: +(1(+1*2^x)/2^x*2^x), so:We get rid of parentheses
1(+1*2^x)/2^x*2^x
We multiply all the terms by the denominator
1(+1*2^x)
Back to the equation:
+(1(+1*2^x))
(1(+1*2^x))-2x)/2^x*2^x=0
We multiply all the terms by the denominator
((1(+1*2^x)))*2^x*2^x-2x)=0
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