log(a)(b)=log(b)(a)(a≠1,b≠1,c≠1),则ab=

题目

log(a)(b)=log(b)(a)(a≠1,b≠1,c≠1),则ab=
（log(x)(y))^2+(log(y)(x))^2=1,则y可用x表示为
前面那个括号是底数

答案

是a≠1,b≠1,a≠b吧?
= = = = = = = = =
1.令 t=log(a)(b),
则 log(b)(a)=1/t.
所以 t =1/t,
解得 t1=1,t2= -1.
又因为 a≠b,
所以 log(a)(b)≠1.
所以 log(a)(b)= -1,
即 b=a^(-1) =1/a.
即 ab=1.
= = = = = = = = =
2.令 t=log(x)(y),x,y属于(0,1)U(1,正无穷大).
则 log(y)(x)=1/t.
即 t^2 +1/(t^2) =1.
即 t^4-t^2+1=0.
该方程无实根.
.
= = = = = = = = =
可能算错了.你再检查一下题目.