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

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



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

We move all terms to the left:

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



Domain of the equation: 3)^(x+1)!=0

x∈R





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

x∈R



We add all the numbers together, and all the variables

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

We calculate fractions


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

We calculate fractions


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

We calculate fractions


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

We can not solve this equation

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