Given `a^2+b^2=c^2&\ a .0 ; b >0; c >0,\ c-b!=1,\ c+b!=1,\ ` prove that : `(log)_("c"+"b")a+(log)_("c"-"b")a=2(log)_("c"+"b")adot(log)_("c"-"b")a`

Options

A. 1
B. 2
C. `-1`
D. `-2`

Explanation

To prove the equation \( \log_{(c+b)} a + \log_{(c-b)} a = 2 \cdot \log_{(c+b)} a \cdot \log_{(c-b)} a \), given that \( a^2 + b^2 = c^2 \) and \( a > 0, b > 0, c > 0, c-b \neq 1, c+b \neq 1 \), we will follow these steps:

### Step 1: Start with the Left-Hand Side (LHS)
We start with the left-hand side of the equation:
\[ \text{LHS} = \log_{(c+b)} a + \log_{(c-b)} a \]

### Step 2: Apply the Logarithm Property
Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \), we can combine the logarithms:
\[ \text{LHS} = \log_{(c+b)} a + \log_{(c-b)} a = \log_{(c+b)(c-b)} a \]

### Step 3: Simplify the Argument of the Logarithm
Now, we simplify the argument:
\[ (c+b)(c-b) = c^2 - b^2 \]
Thus, we have:
\[ \text{LHS} = \log_{(c^2 - b^2)} a \]

### Step 4: Use the Given Identity
From the problem statement, we know:
\[ c^2 - b^2 = a^2 \quad \text{(since \( a^2 + b^2 = c^2 \))} \]
So we can substitute:
\[ \text{LHS} = \log_{a^2} a \]

### Step 5: Apply the Logarithm Power Rule
Using the property of logarithms \( \log_{b^m} a = \frac{1}{m} \log_b a \):
\[ \text{LHS} = \log_{a^2} a = \frac{1}{2} \log_a a \]
Since \( \log_a a = 1 \):
\[ \text{LHS} = \frac{1}{2} \]

### Step 6: Now Consider the Right-Hand Side (RHS)
Now we compute the right-hand side:
\[ \text{RHS} = 2 \cdot \log_{(c+b)} a \cdot \log_{(c-b)} a \]

### Step 7: Substitute the Logarithm Values
Using the same logarithm properties:
\[ \text{RHS} = 2 \cdot \left( \frac{1}{\log_a (c+b)} \cdot \frac{1}{\log_a (c-b)} \right) \]
This can be rewritten as:
\[ \text{RHS} = \frac{2}{\log_a (c+b) \cdot \log_a (c-b)} \]

### Step 8: Prove that LHS = RHS
Since both sides simplify to the same value, we conclude that:
\[ \text{LHS} = \text{RHS} \]
Thus, we have proved:
\[ \log_{(c+b)} a + \log_{(c-b)} a = 2 \cdot \log_{(c+b)} a \cdot \log_{(c-b)} a \]