xn=x⋅x⋅x⋅…⋅x⏟n factors of x x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x...xn=x⋅x⋅x⋅…⋅x⏟n factors of x x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x x} _ {6} =x^ {6}\\ \ dfrac {x^ {5}} {x^ {2}} =\ dfrac {x x x x} {x x} =\ dfrac {\ cancel {(x x)} x x x} {\ cancelar {(x x)}} =x x x=x^ {3}. \ text {Observe que} 5-2=3 Así, por la segunda regla de exponentes,xnxn=xn−n=x0
xn=x⋅x⋅x⋅…⋅x⏟n factors of x x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x...xn=x⋅x⋅x⋅…⋅x⏟n factors of x x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x x} _ {6} =x^ {6}\\ \ dfrac {x^ {5}} {x^ {2}} =\ dfrac {x x x x} {x x} =\ dfrac {\ cancel {(x x)} x x x} {\ cancelar {(x x)}} =x x x=x^ {3}. \ text {Observe que} 5-2=3 Así, por la segunda regla de exponentes,xnxn=xn−n=x0