Find the value of ` (1/49)^(1+log_(7)2) + 5^(-log_((1//5))(7)`.

Explanation

To solve the expression \( \left(\frac{1}{49}\right)^{1 + \log_{7}2} + 5^{-\log_{(1/5)}(7)} \), we will follow these steps:

### Step 1: Rewrite \( \frac{1}{49} \)
We know that \( 49 = 7^2 \), so we can rewrite \( \frac{1}{49} \) as:
\[ \frac{1}{49} = \frac{1}{7^2} = 7^{-2} \]
Thus, we have:
\[ \left(\frac{1}{49}\right)^{1 + \log_{7}2} = (7^{-2})^{1 + \log_{7}2} \]

### Step 2: Apply the power of a power property
Using the property of exponents \( (a^m)^n = a^{m \cdot n} \), we can simplify:
\[ (7^{-2})^{1 + \log_{7}2} = 7^{-2(1 + \log_{7}2)} = 7^{-2 - 2\log_{7}2} \]

### Step 3: Rewrite \( -2\log_{7}2 \)
Using the property \( a \log_{b}c = \log_{b}(c^a) \), we can rewrite \( -2\log_{7}2 \) as:
\[ -2\log_{7}2 = \log_{7}(2^{-2}) = \log_{7}\left(\frac{1}{4}\right) \]
Thus, we have:
\[ 7^{-2 - 2\log_{7}2} = 7^{-2 + \log_{7}\left(\frac{1}{4}\right)} = 7^{\log_{7}\left(\frac{1}{4}\right) / 7^2} \]

### Step 4: Combine the terms
Now, we can write:
\[ 7^{-2 + \log_{7}\left(\frac{1}{4}\right)} = \frac{1}{7^2} \cdot \frac{1}{4} = \frac{1}{49} \cdot \frac{1}{4} = \frac{1}{196} \]

### Step 5: Simplify the second term \( 5^{-\log_{(1/5)}(7)} \)
Using the change of base formula for logarithms, we can rewrite:
\[ -\log_{(1/5)}(7) = -\frac{\log_{10}(7)}{\log_{10}(1/5)} = -\frac{\log_{10}(7)}{-\log_{10}(5)} = \frac{\log_{10}(7)}{\log_{10}(5)} = \log_{5}(7) \]
Thus, we have:
\[ 5^{-\log_{(1/5)}(7)} = 5^{\log_{5}(7)} = 7 \]

### Step 6: Combine both parts
Now we combine both parts:
\[ \frac{1}{196} + 7 \]
To add these, we need a common denominator:
\[ 7 = \frac{7 \cdot 196}{196} = \frac{1372}{196} \]
Thus, we have:
\[ \frac{1}{196} + \frac{1372}{196} = \frac{1373}{196} \]

### Final Answer
The final value of the expression is:
\[ \frac{1373}{196} \]