if `log_10 5=a` and `log_10 3=b` then:

Options

A. ` log_(30) 8 = (3(1-a))/(b+1) `
B. ` log_(40)15= (a+b)/(3-2a)`
C. ` log_(243) 32 = (1-a)/b`
D. none of these

Explanation

To solve the problem, we need to find the logarithmic expressions based on the given values of \( \log_{10} 5 = a \) and \( \log_{10} 3 = b \). We will check each option one by one.

### Step 1: Find \( \log_{30} 8 \)

Using the change of base formula for logarithms:
\[ \log_{30} 8 = \frac{\log_{10} 8}{\log_{10} 30} \]

### Step 2: Express \( \log_{10} 8 \) and \( \log_{10} 30 \)

We can rewrite \( \log_{10} 8 \) as:
\[ \log_{10} 8 = \log_{10} (2^3) = 3 \log_{10} 2 \]
And \( \log_{10} 30 \) can be expressed as:
\[ \log_{10} 30 = \log_{10} (2 \cdot 3 \cdot 5) = \log_{10} 2 + \log_{10} 3 + \log_{10} 5 \]
Substituting the known values:
\[ \log_{10} 30 = \log_{10} 2 + b + a \]

### Step 3: Substitute back into the expression for \( \log_{30} 8 \)

Now substituting \( \log_{10} 8 \) and \( \log_{10} 30 \) into our expression:
\[ \log_{30} 8 = \frac{3 \log_{10} 2}{\log_{10} 2 + b + a} \]

### Step 4: Express \( \log_{10} 2 \)

To express \( \log_{10} 2 \), we can use the fact that:
\[ \log_{10} 2 = \log_{10} \left(\frac{10}{5}\right) = \log_{10} 10 - \log_{10} 5 = 1 - a \]

### Step 5: Substitute \( \log_{10} 2 \) into the expression

Now substituting \( \log_{10} 2 \) into the expression for \( \log_{30} 8 \):
\[ \log_{30} 8 = \frac{3(1 - a)}{(1 - a) + b + a} \]
This simplifies to:
\[ \log_{30} 8 = \frac{3(1 - a)}{1 + b} \]

### Final Result

Thus, we have:
\[ \log_{30} 8 = \frac{3(1 - a)}{1 + b} \]