If x,y,z are in HP, then `log(x+z)+log (x-2y+z)` is equal to

Options

A. `log(x-z)`
B. `2log(x-z)`
C. `3log(x-z)`
D. `4 log(x-z)`

Explanation

To solve the problem, we need to determine the value of \( \log(x+z) + \log(x-2y+z) \) given that \( x, y, z \) are in Harmonic Progression (HP).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression**:
If \( x, y, z \) are in HP, then their reciprocals \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in Arithmetic Progression (AP). This means:
\[ 2y = x + z \]
From this, we can express \( y \):
\[ y = \frac{2xz}{x + z} \]

2. **Substituting for \( y \)**:
We need to find \( x - 2y + z \):
\[ x - 2y + z = x - 2\left(\frac{2xz}{x + z}\right) + z \]
Simplifying this expression:
\[ = x + z - \frac{4xz}{x + z} \]
To combine these terms, we need a common denominator:
\[ = \frac{(x + z)(x + z) - 4xz}{x + z} \]
Expanding the numerator:
\[ = \frac{x^2 + 2xz + z^2 - 4xz}{x + z} = \frac{x^2 - 2xz + z^2}{x + z} \]
This simplifies to:
\[ = \frac{(x - z)^2}{x + z} \]

3. **Finding the Logarithmic Expression**:
Now we substitute back into the logarithmic expression:
\[ \log(x + z) + \log(x - 2y + z) = \log(x + z) + \log\left(\frac{(x - z)^2}{x + z}\right) \]
Using the property of logarithms \( \log a + \log b = \log(ab) \):
\[ = \log\left((x + z) \cdot \frac{(x - z)^2}{x + z}\right) \]
The \( (x + z) \) terms cancel out:
\[ = \log((x - z)^2) \]

4. **Final Simplification**:
Using the logarithmic identity \( \log(a^b) = b \log(a) \):
\[ = 2 \log(x - z) \]

Thus, the final answer is:
\[ \log(x + z) + \log(x - 2y + z) = 2 \log(x - z) \]