If `a^(2)+b^(2)=7ab`, then `log.(1)/(3)(a+b)=(1)/(2)[loga+logb]`

Answer

To prove the equation \( \log \left( \frac{1}{3} (a+b) \right) = \frac{1}{2} \left( \log a + \log b \right) \) given the condition \( a^2 + b^2 = 7ab \), we can follow these steps:

### Step 1: Rewrite the given condition
We start with the equation:
\[ a^2 + b^2 = 7ab \]
We can rearrange this as:
\[ a^2 + b^2 - 7ab = 0 \]

### Step 2: Add \( 2ab \) to both sides
To make it easier to factor, we add \( 2ab \) to both sides:
\[ a^2 + b^2 + 2ab = 7ab + 2ab \]
This simplifies to:
\[ (a + b)^2 = 9ab \]

### Step 3: Take the square root
Taking the square root of both sides gives us:
\[ a + b = 3\sqrt{ab} \]

### Step 4: Substitute into the logarithmic equation
Now we substitute \( a + b \) into the logarithmic equation we want to prove:
\[ \log \left( \frac{1}{3} (a+b) \right) = \log \left( \frac{1}{3} \cdot 3\sqrt{ab} \right) \]
This simplifies to:
\[ \log (\sqrt{ab}) = \frac{1}{2} \log (ab) \]

### Step 5: Use properties of logarithms
Using the property of logarithms that states \( \log (xy) = \log x + \log y \), we can write:
\[ \log (ab) = \log a + \log b \]
Thus, we have:
\[ \log (\sqrt{ab}) = \frac{1}{2} (\log a + \log b) \]

### Step 6: Final substitution
Now we can rewrite our earlier expression:
\[ \log \left( \frac{1}{3} (a+b) \right) = \frac{1}{2} (\log a + \log b) \]
This completes the proof.

### Conclusion
Thus, we have shown that:
\[ \log \left( \frac{1}{3} (a+b) \right) = \frac{1}{2} \left( \log a + \log b \right) \]