Q. If $x,y,z$ are in HP, then $\log{(x + z)} + \log{(x - 2y + z)}$ is equal to
A) $\log{(x - z)}$
B) $2\log{(x - z)}$
C) $3\log{(x - z)}$
D) $4\log{(x - z)}$
Explanation
(b)
Since, $y = \frac{2xz}{x + z}$
Now, $x - 2y + z = x + z - 2\left( \frac{2xz}{x + z} \right)$
$= x + z - \frac{4xz}{x + z} = \frac{(x - z)^{2}}{x + z}$
$\Rightarrow \log{(x - 2y + z)} = \log(x - z)^{2} - \log{(x + z)}$
$\Rightarrow \log{(x - 2y + z)} + \log{(x + z)} = 2\log{(x - z)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.