Respuesta :
remember
x^0=1
also
[tex](x^n)(x^m)=x^{n+m}[/tex]
and
if [tex]a^n=a^m[/tex] where a=a, then assume that n=m
so
[tex](x^3)^0=(x^5)(x^n)[/tex]
1=[tex]x^{5+n}[/tex]
we know that x^0=1
therefor
[tex]x^0=x^{5+n}[/tex]
x^0=x^(5+n) therefor
0=5+n
minus 5 from both sides
-5=n
I can't solve for x since it will be eliminated
n=-5
Joes is correct because of maths
C is answer
x^0=1
also
[tex](x^n)(x^m)=x^{n+m}[/tex]
and
if [tex]a^n=a^m[/tex] where a=a, then assume that n=m
so
[tex](x^3)^0=(x^5)(x^n)[/tex]
1=[tex]x^{5+n}[/tex]
we know that x^0=1
therefor
[tex]x^0=x^{5+n}[/tex]
x^0=x^(5+n) therefor
0=5+n
minus 5 from both sides
-5=n
I can't solve for x since it will be eliminated
n=-5
Joes is correct because of maths
C is answer