If `log_(10)3=x` and `log_(30)5=y`, then `log_(30)8` is equal to

Options

A. `3(1-x-y)`
B. `(1)/(3(1-x-y))`
C. `(3)/((1-x-y))`
D. `(1-x-y)/(3)`

Answer

To solve the problem, we need to find the value of \( \log_{30} 8 \) given that \( \log_{10} 3 = x \) and \( \log_{30} 5 = y \).

### Step-by-Step Solution:

1. **Express \( \log_{30} 8 \) in terms of base 10:**
\[ \log_{30} 8 = \frac{\log_{10} 8}{\log_{10} 30} \]

2. **Rewrite \( \log_{10} 8 \):**
Since \( 8 = 2^3 \), we can use the power rule of logarithms:
\[ \log_{10} 8 = \log_{10} (2^3) = 3 \log_{10} 2 \]

3. **Rewrite \( \log_{10} 30 \):**
Since \( 30 = 3 \times 10 \), we can use the product rule of logarithms:
\[ \log_{10} 30 = \log_{10} (3 \times 10) = \log_{10} 3 + \log_{10} 10 = \log_{10} 3 + 1 \]

4. **Substituting back into the equation for \( \log_{30} 8 \):**
\[ \log_{30} 8 = \frac{3 \log_{10} 2}{\log_{10} 3 + 1} \]

5. **Express \( \log_{10} 2 \) using \( x \) and \( y \):**
We know that:
\[ \log_{10} 3 = x \quad \text{and} \quad \log_{30} 5 = y \]
We can express \( \log_{10} 5 \) in terms of \( y \):
\[ \log_{10} 5 = y \cdot \log_{10} 30 \]
Substituting \( \log_{10} 30 \):
\[ \log_{10} 5 = y \cdot (x + 1) \]

6. **Using the change of base formula to find \( \log_{10} 2 \):**
We know that:
\[ \log_{10} 2 + \log_{10} 3 + \log_{10} 5 = \log_{10} 30 \]
Therefore:
\[ \log_{10} 2 + x + y(x + 1) = x + 1 \]
Simplifying gives:
\[ \log_{10} 2 = 1 - x - y(x + 1) \]

7. **Substituting \( \log_{10} 2 \) back into the equation for \( \log_{30} 8 \):**
\[ \log_{30} 8 = \frac{3(1 - x - y(x + 1))}{x + 1} \]

8. **Final expression for \( \log_{30} 8 \):**
\[ \log_{30} 8 = \frac{3(1 - x - y(x + 1))}{x + 1} \]