If `log_(10)5=a` and `log_(10)3=b`,then

Options

A. `log_(10)8=3(1-a)`
B. `log_(40)15=((a+b))/((3-2a))`
C. `log_(243)32=((1-a)/b)`
D. All of these

Answer

To solve the problem, we need to verify the correctness of the given logarithmic expressions based on the values of \( \log_{10} 5 = a \) and \( \log_{10} 3 = b \).

### Step-by-step Solution:

1. **Option 1: Verify \( \log_{10} 8 = 3(1 - a) \)**

We know:
\[ \log_{10} 8 = \log_{10} (2^3) = 3 \log_{10} 2 \]
Using the change of base formula, we can express \( \log_{10} 2 \) in terms of \( a \) and \( b \):
\[ 1 = \log_{10} 10 = \log_{10} (5 \cdot 2) = \log_{10} 5 + \log_{10} 2 = a + \log_{10} 2 \]
Therefore:
\[ \log_{10} 2 = 1 - a \]
Substituting this into our expression for \( \log_{10} 8 \):
\[ \log_{10} 8 = 3(1 - a) \]
Thus, Option 1 is verified.

2. **Option 2: Verify \( \log_{40} 15 = \frac{a + b}{3 - 2a} \)**

We can express \( \log_{40} 15 \) using the change of base formula:
\[ \log_{40} 15 = \frac{\log_{10} 15}{\log_{10} 40} \]
We know:
\[ \log_{10} 15 = \log_{10} (5 \cdot 3) = \log_{10} 5 + \log_{10} 3 = a + b \]
And:
\[ \log_{10} 40 = \log_{10} (8 \cdot 5) = \log_{10} 8 + \log_{10} 5 = 3(1 - a) + a = 3 - 2a \]
Therefore:
\[ \log_{40} 15 = \frac{a + b}{3 - 2a} \]
Thus, Option 2 is verified.

3. **Option 3: Verify \( \log_{243} 32 = \frac{1 - a}{b} \)**

We can express \( \log_{243} 32 \) using the change of base formula:
\[ \log_{243} 32 = \frac{\log_{10} 32}{\log_{10} 243} \]
We know:
\[ \log_{10} 32 = \log_{10} (2^5) = 5 \log_{10} 2 = 5(1 - a) \]
And:
\[ \log_{10} 243 = \log_{10} (3^5) = 5 \log_{10} 3 = 5b \]
Therefore:
\[ \log_{243} 32 = \frac{5(1 - a)}{5b} = \frac{1 - a}{b} \]
Thus, Option 3 is verified.

4. **Conclusion:**

Since all three options are verified, the answer is:
\[ \text{Option 4: All of these} \]