于是f(x)=9[log(4,x)]^2-[log(2,x)]^2=9[1/2*log(2,x)]^2-[log(2,x)]^2=5/4*[log(2,x)]^2 怎么得的?

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于是f(x)=9[log(4,x)]^2-[log(2,x)]^2=9[1/2*log(2,x)]^2-[log(2,x)]^2=5/4*[log(2,x)]^2 怎么得的?
于是f(x)=9[log(4,x)]^2-[log(2,x)]^2=9[1/2*log(2,x)]^2-[log(2,x)]^2=5/4*[log(2,x)]^2 怎么得的?

于是f(x)=9[log(4,x)]^2-[log(2,x)]^2=9[1/2*log(2,x)]^2-[log(2,x)]^2=5/4*[log(2,x)]^2 怎么得的?
考虑换底公式.log(4,x)=lgx/lg4=lgx/(2lg2)=1/2*lgx/lg2=1/2*log(2.x)
故f(x)=9[log(4,x)]^2-[log(2,x)]^2=9[1/2*log(2,x)]^2-[log(2,x)]^2=(9/4-1)[log(2,x)]^2=5/4*[log(2,x)]^2