若a>b>1且loga,b+logb,aa=二份之五,求loga,b-logb,a的值.

Question

若a>b>1且log a ,b+logb,a a=二份之五,求log a,b-log b ,a 的值.

Answer 1

换底公式
lgb/lga+lga/lgb=5/2
令x=lgb/lga
则x+1/x=5/2
2x²-5x+2=0
(2x-1)(x-2)=0
a>b>1
所以lga>lgb>0
所以lgb/lga<1
即x<1
所以x=1/2
所以log a,b-log b ,a =lgb/lga-lga/lgb=x-1/x=1/2-2=-3/2