Answer :
so solving for x
minus c1 from both sides
x^n=c2-c1
take the nth root of both sides
[tex]x=\sqrt[n]{c_2-c_1}[/tex]
dunno about the relationship between them but that's what x is
minus c1 from both sides
x^n=c2-c1
take the nth root of both sides
[tex]x=\sqrt[n]{c_2-c_1}[/tex]
dunno about the relationship between them but that's what x is