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

icaptain 1年前 已收到2个回答 举报

郡儿公主 花朵

共回答了23个问题采纳率:87% 举报

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

1年前

4

高盛 幼苗

共回答了6个问题 举报

1/(log以3x为底3的对数)=log以3为底3x的对数=1+log以3为底x的对数
接下来解方程你应该会了

1年前

0
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