题目

解方程1/（log以3x为底3的对数） 加上 (log以3为底x的对数)^2=3

提问时间：2020-08-01

答案

1/（log以3x为底3的对数） 加上 (log以3为底x的对数)^2=3
3x>0,3x≠1,x>0
[1/(log(3x)(3))]+[log(3)(x)]^2=3
又因为[1=log(3x)(3x)]
所以[log(3x)(3x)/log(3x)(3)]+[log(3)(x)]^2=3
又因为[log(3x)(3x)/log(3x)(3)=log(3)(3x)]即{log(c)(b)/log(c)(a)=log(a)(b)}
所以log(3)(3*x)+[log(3)(x)]^2=3
又因为[log(3)(3*x)=log(3)(3)+log(3)(x)]即{log(a)(b*c)=log(a)(b)+log(a)(c)}
所以[log(3)(3)]+[log(3)(x)]+[log(3)(x)]^2=3
所以1+log(3)(x)+[log(3)(x)]^2=3
所以log(3)(x)+[log(3)(x)]^2=2
所以[log(3)(x)+1/2]^2-1/4=2
所以[log(3)(x)+1/2]^2=9/4
所以log(3)(x)+1/2=±3/2
所以log(3)(x)=-1/2±3/2
所以x=3^(-1/2±3/2)
x=3^1=3或x=3^(-2)=1/9
x=3或x=1/9
以下,
[1/(log(3x)(3))]+[log(3)(x)]^2=3
[log(3x)(3x)/log(3x)(3)]+[log(3)(x)]^2=3
log(3)(3*x)+[log(3)(x)]^2=3
[log(3)(3)]+[log(3)(x)]+[log(3)(x)]^2=3
1+log(3)(x)+[log(3)(x)]^2=3
[log(3)(x)+1/2]^2=9/4
log(3)(x)=-1/2±3/2
x=3^1=3或x=3^(-2)=1/9
x=3或x=1/9