If `log(x+z)+log(x-2y+z)=2log(x-z)," then "x,y,z` are in

Options

A. H.P.
B. G.P.
C. A.P.
D. none of these

Answer

To solve the equation \( \log(x+z) + \log(x-2y+z) = 2\log(x-z) \) and determine the relationship between \( x, y, z \), we can follow these steps:

### Step 1: Use Logarithmic Properties
We start with the given equation:
\[ \log(x+z) + \log(x-2y+z) = 2\log(x-z) \]
Using the property of logarithms that states \( \log a + \log b = \log(ab) \), we can rewrite the left-hand side:
\[ \log((x+z)(x-2y+z)) = \log((x-z)^2) \]

### Step 2: Remove the Logarithm
Since the logarithm function is one-to-one, we can equate the arguments:
\[ (x+z)(x-2y+z) = (x-z)^2 \]

### Step 3: Expand Both Sides
Now, we will expand both sides of the equation:
- Left-hand side:
\[ x^2 - 2xy + xz + xz + z^2 = x^2 - 2xy + 2xz + z^2 \]
- Right-hand side:
\[ (x-z)(x-z) = x^2 - 2xz + z^2 \]

### Step 4: Set the Equations Equal
Now we have:
\[ x^2 - 2xy + 2xz + z^2 = x^2 - 2xz + z^2 \]

### Step 5: Simplify the Equation
Subtract \( x^2 + z^2 \) from both sides:
\[ -2xy + 2xz = -2xz \]
Now, add \( 2xz \) to both sides:
\[ -2xy + 4xz = 0 \]

### Step 6: Factor Out Common Terms
Factoring out \( 2 \):
\[ 2(-xy + 2xz) = 0 \]
This gives us:
\[ -xy + 2xz = 0 \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ xy = 2xz \]

### Step 8: Divide by \( xz \) (assuming \( x \neq 0 \) and \( z \neq 0 \))
Dividing both sides by \( xz \):
\[ \frac{y}{z} = 2 \]
This implies:
\[ y = 2z \]

### Conclusion
The relationship derived indicates that \( x, y, z \) are in Harmonic Progression (HP).