If `a^(2) + b^(2) = 7ab`, then the vlaue of `log [(1)/(3) (a + b)]` is

Options

A. `(1)/(2) log a log b`
B. `(1)/(2) log ab`
C. `(1)/(2) [log a + log b]`
D. `(1)/(2) [log a - log b]`

Answer

To solve the problem, we start with the given equation:

**Step 1: Rewrite the equation**
Given:
\[ a^2 + b^2 = 7ab \]

**Step 2: Use the identity for \( (a + b)^2 \)**
We know that:
\[ (a + b)^2 = a^2 + b^2 + 2ab \]
Substituting the value of \( a^2 + b^2 \):
\[ (a + b)^2 = 7ab + 2ab = 9ab \]

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

**Step 4: Find \( \frac{1}{3}(a + b) \)**
Now, we calculate:
\[ \frac{1}{3}(a + b) = \frac{1}{3}(3\sqrt{ab}) = \sqrt{ab} \]

**Step 5: Take the logarithm**
We need to find:
\[ \log\left(\frac{1}{3}(a + b)\right) = \log(\sqrt{ab}) \]

Using the properties of logarithms:
\[ \log(\sqrt{ab}) = \frac{1}{2} \log(ab) \]

**Step 6: Expand \( \log(ab) \)**
Using the property of logarithms:
\[ \log(ab) = \log(a) + \log(b) \]

Thus,
\[ \log\left(\frac{1}{3}(a + b)\right) = \frac{1}{2}(\log(a) + \log(b)) \]

**Final Result:**
The value of \( \log\left(\frac{1}{3}(a + b)\right) \) is:
\[ \frac{1}{2}(\log(a) + \log(b)) \]

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