# 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

If the positive numbers x,y,z are in harmonic progression, then $\mathrm{log}\left(x+z\right)+\mathrm{log}\left(x-2y+z\right)$ equals
$a\right)4\mathrm{log}\left(x-z\right)\phantom{\rule{0ex}{0ex}}b\right)3\mathrm{log}\left(x-z\right)\phantom{\rule{0ex}{0ex}}c\right)2\mathrm{log}\left(x-z\right)\phantom{\rule{0ex}{0ex}}d\right)\mathrm{log}\left(x-z\right)$
How do i approach this problem? IF x,y,z are in HP, $⇒y=2xy/x+z$
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Remington Wells
We need to eliminate $y$
$x+z-2y=x+z-4zx/\left(z+x\right)=\frac{\left(z-x{\right)}^{2}}{z+x}$
$\mathrm{log}\left(z+x\right)+\mathrm{log}\left(x+z-2y\right)=\mathrm{log}\left(z+x\right)\left(x+z-2y\right)$
$=\mathrm{log}\left(z+x\right)\cdot \frac{\left(z-x{\right)}^{2}}{z+x}=\mathrm{log}\left(z-x{\right)}^{2}$
If $x>z,\mathrm{log}\left(z-x{\right)}^{2}=2\mathrm{log}\left(x-z\right)$