设a>b>1,log(a)b+log(b)a=10/3,则log(a)b-log(b)a=

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设a>b>1,log(a)b+log(b)a=10/3,则log(a)b-log(b)a=
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设a>b>1,log(a)b+log(b)a=10/3,则log(a)b-log(b)a=
设a>b>1,log(a)b+log(b)a=10/3,则log(a)b-log(b)a=

设a>b>1,log(a)b+log(b)a=10/3,则log(a)b-log(b)a=
即lgb/lga+lga/lgb=10/3
令x=lgb/lga
则x+1/x=10/3
3x²-10x+3=0
x=3,x=1/3
因为a>b>1
所以0

因为logab=lgb/lga logba=lga/lgb
于是令logab=lgb/lga =t
则3t+3/t=10
解得t=3或t=1/3(又a>b>1,故logab>1)
即t=3
因此log(a)(b)-log(b)(a)=t-1/t=3-1/3=8/3.