If x,y,z are in H.P,prove that log(x+z)+log(x+z-2y)=2log(x-z).
106x,y,z are in HP
==> 1/x,1/y,1/z is in AP
==> yz,xz,xy is in AP (multiplying by abc)
==> 2xz = yz+xy .. (i)
Now, log(x+z)+log(x+z-2y)
= log(x+z)(x+z-2y)
= log(x2+xz-2xy+xz+z2-2yz)
= log(x2+y2 - 2xz) [2xz-2xy-2yz = 2xz - 4xz = -2xz from (i)]
= log(x-z)2
= 2log(x-z)
1we have to prove log(x+z)+log(x+z-2y)=2log(x-z).
using property of log
log(x+z)(x+z-2y)=log(x-z)2
taking antilog
x2+xz-2xy+xz+z2-2yz=x2+z2-2xz
2xz-2xy-2yz=-2xz
4xz=2xy+2yz
dividing both sides by xyz we have
2/y=1/x+1/z
which is given as x,y &z are in h.p.
