Question

已知:a的m次幂=3,a的n次幂=2,求(a的3m+n次幂)的平方,

Answer 1

a^m=3,a^n=2,

[a^(3m+n)]^2

=[a^(3m)*a^n)]^2

=[(a^m)^3 *a^n)]^2

=[(3)^3 *2)]^2

=[(3)^3 *2)]^2

=3^6 * 2^2

=729*4

=2916