If the positive numbers x,y,z are in harmonic progression, then log(x+z) + log(x-2y+z) equals

Question

If the positive numbers x,y,z are in harmonic progression, then log(x+z) + log(x-2y+z) equals

a) 4log(x-z) b) 3log(x-z) c) 2log(x-z) d) log(x-z)

How do i approach this problem? IF x,y,z are in HP, => y=2xy/x+z

Answer 1

We need to eliminate $y$

$x+z-2y=x+z-4zx/(z+x)=\dfrac{(z-x)^2}{z+x}$

$\displaystyle\log(z+x)+\log(x+z-2y)=\log(z+x)(x+z-2y)$ $\displaystyle=\log(z+x)\cdot\dfrac{(z-x)^2}{z+x}=\log(z-x)^2$

If $x>z,\log(z-x)^2=2\log(x-z)$