Question

Solve if x=log 6(base 12) y=log 12(base 18) and z=log18(base24),prove that 1+xyz=2yz.

Answer 1

1 + xyz = 1 + log 12 16 x log 18 12 x log 24 18

= 1 + log 6/log 12 x log 12/log 18 x log 18/log 24 = 1 + log 6/log 24

= log 24 24  + log 24 6

= log 24 ( 6 x 24) = log 24 (12x12)

Now 2yz = 2 x log 18 12 x log 24 18 = 2 x log 12/log 18 x log 18 / log 24 = 2 log base 24 12

= log base 24(12^2) = log base 24 (12x12)

Hence 1 + xyz = 2yz

Answer 2

solution:

Your question is wrong, it should be,

X = log12(base 6) , y = log18(base 12) and z = log24( base 18 )

So, 1+xyz = 1 + log12(base 6 )+ og18( base 12) + log24( base 18)

                 = 1+ log6/log12 x log12/log18 x log18/log24

   = log24( base 24) =log24( base 6 )

   = log24(6 x 24)

   = log24 (12 x 12)

Now,  

2yz = 2 x log18(base 12) x log24( base18 )

      =  2 x log12/log18 x log18/log24

     =  2log24( base12)

      = log24(12(base2) )

      = log24(12 x 12)

Hence, 1+xyz = 2yz