solve
`|[1,log_(a)b,log_(a)c],[log_(b)a,1,log_(b)c],[log_(c)a,log_(c)b,1]|=0`

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

(1+log_(c)a)log_(a)x*log_(b)c=log_(b)x log_(a)x

a^(log_(b)c)=c^(log_(b)a)

4.Prove that log_(a)(bc).log_(b)(ca).log_(c)(ab)=2+log_(a)(bc)+log_(b)(ca)+log_(c)(ab)

If a,b,c are distinct real number different from 1 such that (log_(b)a. log_(c)a-log_(a)a) + (log_(a)b.log_(c)b.log_(c)b-log_(b)b) +(log_(a)c.log_(b)c-log_(c)C)=0 , then abc is equal to

Let a,b" and "c are distinct positive numbers,none of them is equal to unity such that log _(b)a .log_(c)a+log_(a)b*log_(c)b+log_(a)c*log_(b)c-log_(b)a sqrt(a)*log_(sqrt(c))b^(1/3)*log_(a)c^(3)=0, then the value of abc is -

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

If a,b,c are distinct positive real numbers each different from unity such that (log_(a)a.log_(c)a-log_(a)a)+(log_(a)b*log_(c)b-log b_(b))+(log_(a)c.log_(a)c-log_(c)c)=0 then prove that abc=1