Let M denote antilog `""_(32)`0.6 and N denote the value of `49^((1-log_(7)2))+5^(-log_(5)4)`. Then M.N is :

Options

A. 100
B. 400
C. 50
D. 200

Answer

To solve the given problem, we will compute the values of \( M \) and \( N \) step by step and then find \( M \cdot N \).

### Step 1: Calculate \( M \)

We are given:
\[ M = \text{antilog}_{32}(0.6) \]

Using the property of antilogarithm, we can express this as:
\[ M = 32^{0.6} \]

Next, we can rewrite \( 32 \) as a power of \( 2 \):
\[ 32 = 2^5 \]
Thus,
\[ M = (2^5)^{0.6} = 2^{5 \cdot 0.6} = 2^{3} = 8 \]

### Step 2: Calculate \( N \)

We are given:
\[ N = 49^{(1 - \log_{7}(2))} + 5^{(-\log_{5}(4))} \]

First, we can simplify \( 49 \):
\[ 49 = 7^2 \]
So,
\[ N = (7^2)^{(1 - \log_{7}(2))} + 5^{(-\log_{5}(4))} \]

Using the property of exponents:
\[ N = 7^{2(1 - \log_{7}(2))} + 5^{-\log_{5}(4)} \]

Now, simplify \( 7^{2(1 - \log_{7}(2))} \):
\[ N = 7^{2 - 2\log_{7}(2)} = 7^2 \cdot 7^{-2\log_{7}(2)} = 49 \cdot 2^{-2} = 49 \cdot \frac{1}{4} = \frac{49}{4} \]

Next, simplify \( 5^{-\log_{5}(4)} \):
\[ 5^{-\log_{5}(4)} = \frac{1}{5^{\log_{5}(4)}} = \frac{1}{4} \]

Now, combine both parts:
\[ N = \frac{49}{4} + \frac{1}{4} = \frac{49 + 1}{4} = \frac{50}{4} = \frac{25}{2} \]

### Step 3: Calculate \( M \cdot N \)

Now that we have both \( M \) and \( N \):
\[ M \cdot N = 8 \cdot \frac{25}{2} \]
Calculating this gives:
\[ M \cdot N = 4 \cdot 25 = 100 \]

### Final Answer
Thus, the value of \( M \cdot N \) is:
\[ \boxed{100} \]