Find the value of `49^((1-log_7(2)))+5^(-log_5(4)` is

Options

A. 10.5
B. 11.5
C. 12.5
D. 13.5

Answer

To find the value of \( 49^{(1 - \log_7(2))} + 5^{(-\log_5(4))} \), we can follow these steps:

### Step 1: Rewrite the bases
First, we can rewrite \( 49 \) as \( 7^2 \):
\[ 49^{(1 - \log_7(2))} = (7^2)^{(1 - \log_7(2))} \]

### Step 2: Apply the power of a power property
Using the property of exponents \( (a^m)^n = a^{m \cdot n} \):
\[ (7^2)^{(1 - \log_7(2))} = 7^{2(1 - \log_7(2))} \]

### Step 3: Distribute the exponent
Distributing the exponent gives:
\[ 7^{2(1 - \log_7(2))} = 7^{2 - 2\log_7(2)} \]

### Step 4: Use the logarithmic identity
Using the identity \( a^{\log_a(b)} = b \), we can rewrite \( 2\log_7(2) \) as \( \log_7(2^2) = \log_7(4) \):
\[ 7^{2 - 2\log_7(2)} = \frac{7^2}{7^{\log_7(4)}} = \frac{49}{4} \]

### Step 5: Simplify the second term
Now, let's simplify the second term \( 5^{(-\log_5(4))} \):
Using the identity \( a^{-\log_a(b)} = \frac{1}{b} \):
\[ 5^{(-\log_5(4))} = \frac{1}{4} \]

### Step 6: Combine the results
Now we can combine both parts:
\[ 49^{(1 - \log_7(2))} + 5^{(-\log_5(4))} = \frac{49}{4} + \frac{1}{4} = \frac{49 + 1}{4} = \frac{50}{4} = \frac{25}{2} \]

### Final Answer
Thus, the final value is:
\[ \frac{25}{2} \]

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