If `log_(2)m=sqrt(7)andlog_(7)n=sqrt(2)`,mn=

Options

A. 1
B. 2
C. 96
D. 98

Explanation

To solve the problem, we need to find the product \( mn \) given the equations \( \log_2 m = \sqrt{7} \) and \( \log_7 n = \sqrt{2} \).

### Step 1: Convert logarithmic equations to exponential form
From the first equation \( \log_2 m = \sqrt{7} \), we can convert it to exponential form:
\[ m = 2^{\sqrt{7}} \]
From the second equation \( \log_7 n = \sqrt{2} \), we can also convert it to exponential form:
\[ n = 7^{\sqrt{2}} \]

### Step 2: Find the product \( mn \)
Now, we can find the product \( mn \):
\[ mn = m \cdot n = (2^{\sqrt{7}}) \cdot (7^{\sqrt{2}}) \]

### Step 3: Simplify the expression
We can write the product as:
\[ mn = 2^{\sqrt{7}} \cdot 7^{\sqrt{2}} \]

### Step 4: Calculate the numerical value
To find the numerical value of \( mn \), we can calculate \( 2^{\sqrt{7}} \) and \( 7^{\sqrt{2}} \):
1. Calculate \( \sqrt{7} \approx 2.64575 \), so \( 2^{\sqrt{7}} \approx 2^{2.64575} \approx 6.2582 \).
2. Calculate \( \sqrt{2} \approx 1.41421 \), so \( 7^{\sqrt{2}} \approx 7^{1.41421} \approx 15.673 \).

Now, multiply these two results:
\[ mn \approx 6.2582 \cdot 15.673 \approx 98.0 \]

### Final Answer
Thus, the product \( mn \) is approximately \( 98 \).

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