5*|[1,log_(b)a],[log_(a)b,1]|

If a,b,c gt1 then Delta=|(log_(a)(abc),log_(a)b,log_(a)c),(log_b(abc),1,log_(b)c),(log_(c)(abc),log_(c)b,1)| equals

If a>1,b>1, then the minimum value of log_(b)a+log_(a)b is

(1)/(1+log_(b)a+log_(b)c)+(1)/(1+log_(c)a+log_(c)b)+(1)/(1+log_(a)b+log_(a)c)

( Prove that )/(1+log_(b)a+log_(b)c)+(1)/(1+log_(c)a+log_(c)b)+(1)/(1+log_(a)b+log_(a)c)=1

(1)/(1+log_(b)a+log_(b)c)+(1)/(1+log_(c)a+log_(c)b)+(1)/(1+log_(a)b+log_(a)c) has the value equal to

Show that log_(b)a log_(c)b log_(a)c=1

If agt1,bgt1,cgt1,dgt1 then the minimum value of log_(b)a+log_(a)b+log_(d)c+log_(c)d , is

Let ABC be a triangle right at C. The value (log_(b+c)a+log_(c-b)a)/(log_(b+c)a*log_(c-b)a)(b+c!=1,c-b!=1) equals

[[(log_a^b),1],[1,log_b^a]] =................ .