Do you mean "if a(b+c-a)/loga=b(c+a-b)/logb=c(a+b-c)/logc show that b^c×c^b=c^a×a^c"?
Let r=a(b+c-a)/log(a) and therefore the other two expressions.
log(ar)=a(b+c-a), log(br)=b(c+a-b), log(cr)=c(a+b-c).
log(a)=a(b+c-a)/r, log(b)=b(c+a-b)/r, log(c)=c(a+b-c)/r.
log(b)-log(a)=(a-b)(a+b-c)/r=(a-b)log(c)/c. Therefore c(log(b)-log(a))=(a-b)log(c) (1)
If bccb=caac, clog(b)+blog(c)=alog(c)+clog(a), c(log(b)-log(a))=(a-b)log(c) (2)
(1) and (2) are identical, so bccb=caac⇒c(log(b)-log(a))=(a-b)log(c).