已知函数f(X)=1+loga X {以a为底}(a>0,a不等于1),满足f(9)=3,则f-1{反函数}(log以9为底2)的值

问题描述:

已知函数f(X)=1+loga X {以a为底}(a>0,a不等于1),满足f(9)=3,则f-1{反函数}(log以9为底2)的值

原函数中f(9)=3
∴1+log(a底)9=3
∴log(a底)9=2
∴a²=9,a=3
∴f(x)=1+log(3底)x
求反函数值f-1(log以9为底2)
即在原函数中y=log(9底)2,求x
由1+log(3底)x=log(9底)2
log(3底)x=1/2log(3底)2 -1
log(3底)x=log(3底)(√2 /3)
∴x=√2/3
∴f-1{反函数}(log以9为底2)=√2/3