Question

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. $4\log (x-z)$

Answer 1

The correct option is B

$2\log (x-z)$

Explanation for the correct answer:

Simplification of logarithm expression:

As $x,y,z$ are in harmonic progression, $y$ is the harmonic mean of $x$ and $z$

$\Rightarrow$ $y=\frac{2xz}{x+z}$ $...\left(i\right)$

To find the value of the required expression, we substitute the value of $y$ from $\left(i\right)$

$\Rightarrow$ $\log (x+z)+\log (x-2y+z)$$=\log (x+z)+\log \left(x-2\times \frac{2xz}{x+z}+z\right)$

$=\log (x+z)+\log \left(\frac{x\left(x+z\right)-4xz+\left(x+z\right)z}{x+z}\right)$

$=\log (x+z)+\log \left(\frac{x^2-2xz+z^2}{x+z}\right)$

$=\log (x+z)+\log \left(\frac{{\left(x-z\right)}^2}{x+z}\right)$

$=\log \left(\frac{\left(x+z\right){\left(x-z\right)}^2}{x+z}\right)$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \log a+\log b=\log \mathrm{ab}\right]$

$=\log \left({\left(x-z\right)}^2\right)$

$\therefore$$\log (x+z)+\log (x-2y+z)$ $=2\log \left(x-z\right)$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \log a^m=m\times \log a\right]$

Hence, the value of the expression $\log (x+z)+\log (x-2y+z)$ is $2\log (x-z),$

Hence, option (B) is the correct answer.