Question

.

Answer 1

If x = 1+ log a^bc, y=1+log b^ca and z=1+log c^ab, prove that xyz = xy+yz+zx.

Answer 2

= 1+ log a bc =loga a+ loga bc
= loga abc
therefore 1/x=logabc a
similarly y = log b abc
z= log c abc
(xy +yz+ zx)/xyz
= 1/z + 1/x +1/y


= logabc c +logabc a +logabc b
= log abc abc
=1
(xy +yz+ zx)/xyz = 1
therefore xy + yz +zx =xyz

Answer 3

http://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.67011.html

Answer 4

Yes i saw that earlier i didn't get it

Answer 5

I didn't get that sollution but

Answer 6

Oh I'm sorry

Answer 7

@hartnn

Answer 8

@sidsiddhartha

Answer 9

use this property