log[x]y=log[y]x

a) show that log[y]x=1/log[x]y
LHS=log[y]x
=(logx)/(logy)
=1/[(logy)/(logx)]
=1/log[x]y
b) given that log[x]y=log[y]x, show that y=1/x
by (a),(log[x]y)^2=1
log[x]y=1 or -1 AND log[y]x=1 or -1
logx=logy or logx=-logy
x=y(rejected as x>y) or x=y^-1
y=1/x
(c)log[x]y=log[x]y...................1
log[x](x-y)=log[y](x+y)............2
from (1),y=1/x ......................3 (by( b))
put(3)into(2)
[log(x-1/x)]/logx=[log(x+1/x)]/(-logx)
-log(x-1/x)=log(x+1/x)
1/(x-1/x)=x+1/x
x^2-(1/x)^2=1
x^4-x^2-1
x^2=(1+(5)^(1/2))/2 or (1-(5)^(1/2))/2(rejected x^2>0)
x=1.27 or-1.27(rejected as x>0)
x=1.27 y=0.786

2010-01-23 09:33:05 補充：
x>y means x is greater than y
x>0 means x is greater than 0

I think you and me make a mistake when typing
log[x]y=log[x]y should be log[x]y=log[y]x