已知3^m=6,9^n=2,求3^(2m-4n+1)
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已知3^m=6,9^n=2,求3^(2m-4n+1)
已知3^m=6,9^n=2,求3^(2m-4n+1)
已知3^m=6,9^n=2,求3^(2m-4n+1)
由3^m=6,9^n=2得
m=log(3)6;(底数为3,真数为6的对数)
n=log(9)2=(1/2)·log(3)2(底数为3,真数为2的对数)
那么,2m-4n+1
=2·log(3)6-2·log(3)2+1
=2·[log(3)6-log(3)2]+1
=2·log(3)3+1
=3
因此,3^(2m-4n+1)
=3^3
=27
9^n=3^2n=2
3^m/3^2n=6/2=3
m-2n=1
2m-4n+1=3
3^(2m-4n+1)=27
^m =6;9^n =2,则
3^(2m-4n+1)=(3^m)^2/(3^2n)^2*3=6^2/2^2*3=27
3^(2m-4n+1)=(3^m)^2/(3^2n)^2*3
因为3^m=6,9^n=2
所以原式=6^2/2^2*3
=27
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