The number `N = 6 log_(10) 2+ log_(10) 31 ` lies between two successive integers whose sum is equal to

Options

A. 5
B. 7
C. 9
D. 10

Answer

To find the value of \( N = 6 \log_{10} 2 + \log_{10} 31 \) and determine between which two successive integers it lies, we can follow these steps:

### Step 1: Calculate \( \log_{10} 2 \)
We know that \( \log_{10} 2 \approx 0.301 \).

### Step 2: Calculate \( \log_{10} 31 \)
We can also approximate \( \log_{10} 31 \). A common approximation is \( \log_{10} 31 \approx 1.491 \).

### Step 3: Substitute the values into the equation for \( N \)
Now, substitute these values into the equation:
\[ N = 6 \log_{10} 2 + \log_{10} 31 \]
\[ N = 6 \times 0.301 + 1.491 \]

### Step 4: Calculate \( 6 \times 0.301 \)
Calculating this gives:
\[ 6 \times 0.301 = 1.806 \]

### Step 5: Add \( 1.806 \) and \( 1.491 \)
Now, add these two results:
\[ N = 1.806 + 1.491 = 3.297 \]

### Step 6: Identify the two successive integers
The value \( 3.297 \) lies between the integers \( 3 \) and \( 4 \).

### Step 7: Calculate the sum of these integers
Now, we find the sum of these two integers:
\[ 3 + 4 = 7 \]

### Conclusion
Thus, the number \( N \) lies between the integers \( 3 \) and \( 4 \), and their sum is \( 7 \).