If `log_(a)5=x and log_(a)7=y`, then `log_(a)sqrt(1,4)=`

Options

A. `(1)/(2)xy`
B. `(1)/(2)x-y`
C. `(1)/(2)(x+y)`
D. `(1)/(2)(y-x)`

Explanation

To solve the problem, we need to find \( \log_a \sqrt{1.4} \) given that \( \log_a 5 = x \) and \( \log_a 7 = y \).

### Step-by-Step Solution:

1. **Rewrite the logarithm of the square root**:
\[ \log_a \sqrt{1.4} = \log_a (1.4)^{1/2} \]
This uses the property that the logarithm of a power can be expressed as the exponent times the logarithm of the base.

**Hint**: Remember that \( \sqrt{a} = a^{1/2} \).

2. **Apply the power rule of logarithms**:
\[ \log_a (1.4)^{1/2} = \frac{1}{2} \log_a (1.4) \]

**Hint**: The power rule states that \( \log_b (m^n) = n \cdot \log_b (m) \).

3. **Express 1.4 as a fraction**:
\[ 1.4 = \frac{7}{5} \]

**Hint**: Knowing that \( 1.4 \) can be expressed as a fraction will help simplify the logarithm.

4. **Substitute into the logarithm**:
\[ \log_a (1.4) = \log_a \left(\frac{7}{5}\right) \]

**Hint**: Use the property of logarithms that states \( \log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n \).

5. **Apply the quotient rule of logarithms**:
\[ \log_a \left(\frac{7}{5}\right) = \log_a 7 - \log_a 5 \]

**Hint**: This step is crucial as it breaks down the logarithm into manageable parts.

6. **Substitute the values of \( \log_a 7 \) and \( \log_a 5 \)**:
\[ \log_a 7 = y \quad \text{and} \quad \log_a 5 = x \]
Therefore,
\[ \log_a \left(\frac{7}{5}\right) = y - x \]

**Hint**: Substitute the known values directly to simplify your expression.

7. **Combine everything**:
\[ \log_a \sqrt{1.4} = \frac{1}{2} \log_a \left(\frac{7}{5}\right) = \frac{1}{2} (y - x) \]

**Hint**: Make sure to keep track of the coefficients when combining terms.

8. **Final answer**:
\[ \log_a \sqrt{1.4} = \frac{1}{2} (y - x) \]

Thus, the answer is \( \frac{1}{2} y - \frac{1}{2} x \), which corresponds to option D.

### Summary of the Solution:
The final result is:
\[ \log_a \sqrt{1.4} = \frac{1}{2} (y - x) \]