(9^x+1)+(3^2x+1)=729

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Solution for (9^x+1)+(3^2x+1)=729 equation: 



(9^x+1)+(3^2x+1)=729

We move all terms to the left:

(9^x+1)+(3^2x+1)-(729)=0

We get rid of parentheses

9^x+3^2x+1+1-729=0

We add all the numbers together, and all the variables

9^x+3^2x-727=0

We move all terms containing x to the left, all other terms to the right

9^x+3^2x=727

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