32(1/7)=2^5x+1

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Solution for 32(1/7)=2^5x+1 equation: 



32(1/7)=2^5x+1

We move all terms to the left:

32(1/7)-(2^5x+1)=0

We add all the numbers together, and all the variables

-(2^5x+1)+32(+1/7)=0

We multiply parentheses

-(2^5x+1)+1/7*32=0

We get rid of parentheses

-2^5x-1+1/7*32=0

We multiply all the terms by the denominator


-2^5x*7*32+1-1*7*32=0

We add all the numbers together, and all the variables

-2^5x*7*32-223=0

Wy multiply elements

-448x^5*3-223=0

We do not support expression: x^5

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