题目

公式变形
1:已知a^m=2,a^n=6求a^2m+3n
2:已知a^x+y=6,a^x+y+z=12求a^z
3:已知a^m=2 b^m=6求(a^3b^2)^m的值
4已知5^m=3,5^3m+n+1=270求5^n

提问时间：2020-07-13

答案

1:已知a^m=2,a^n=6求a^2m+3n
a^2m+3n
=a^2m*a^3n
=(a^m)^2*(a^n)^3
=2^2*6^3
=4*216
=864
2:已知a^x+y=6,a^x+y+z=12求a^z
a^x+y+z
=a^(x+y)*a^z
=12
因为a^x+y=6,所以a^z=12/6=2
3:已知a^m=2 b^m=6求(a^3b^2)^m的值
(a^3b^2)^m
=a^3m*b^2m
=(a^m)3*(b^m)^2
=2^3*6^2
=8*36
=288
4已知5^m=3,5^3m+n+1=270求5^n
5^3m+n+1
=5^3m*5^n*5
=(5^m)^3*5^n*5
=3^3*5^n*5
=27*5^n*5
=270
所以5^n=270/(27*5)=2