Question

find n.

x^8 x x^n = x^20

Answer 1

\[\Large x^8 \times x^n = x^{20}\]


\[\Large x^{8+n}= x^{20}\]


Since the bases are equal, the exponents are equal


So 8 + n = 20


8 + n = 20


8 + n -8 = 20-8


n = 12

Answer 2

thanks! could you help me with x^n x x^-3 =x^8

Answer 3

sure thing


\[\Large x^n \times x^{-3} = x^8\]


\[\Large x^{n+(-3)} = x^8\]


Since the bases are equal, the exponents are equal, so n+(-3) = 8


n+(-3) = 8


n-3 = 8


n-3+3 = 8 + 3


n = 11