`a`, `b`, `c` are positive integers formaing an incresing `G.P.` and `b-a` is a perfect cube and `log_(6)a+log_(6)b+log_(6)c=6`, then `a+b+c=`

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

a, b, c are positive integers froming an increasing G.P . whose common ratio is rational number, b - a is cube of natural number and log_(6)a + log_(6)b + log_(6)c = 6 then a + b + c is divisible by

If a, b, c, are positive real numbers and log_(4) a=log_(6)b=log_(9) (a+b) , then b/a equals

If a,b,c are in GP then log_(10)a, log_(10)b, log_(10)c are in

If log_(4)A=log_(6)B=log_(9)(A+B), then [4((B)/(A))]

If a=log_(4)5 and b=log_(5)6, then log_(2)3=

If a,b and c are positive numbers in a GP,then the roots of the quadratic equation (log_(e)a)^(2)-(2log_(e)b)x+(log_(e)c)=0 are

If a=log_(4)5 and b=log_(5)6 then log_(2)3 is