To evaluate \( \log(104 \cdot 103) \), we can use the logarithmic property that states:
\[ \log(a \cdot b) = \log(a) + \log(b) \]
So, we have:
\[ \log(104 \cdot 103) = \log(104) + \log(103) \]
Now, we can break down \( 104 \) and \( 103 \):
\[ 104 = 10^2 + 4 = 100 + 4 = 10^2 + 4 \text{ (however, we'll keep it as is for simplicity)} \] \[ 103 = 10^2 + 3 \text{ (as with 104)} \]
However, it looks like these numbers don't lend themselves to simplicity in direct logarithmic calculations without a calculator.
Given the choices available:
- \( 12 \)
- \( \log 4 + 3 \)
- \( 7 \)
- \( 4 + \log 3 \)
Let's estimate the answer. We know:
\[ 104 \approx 100 \quad \text{and} \quad 103 \approx 100 \] So, \[ \log(104) \approx 2 \quad \text{and} \quad \log(103) \approx 2 \] Thus, \[ \log(104 \cdot 103) \approx 2 + 2 = 4 \]
This is not on the list, so we employ a different approach:
\( \log(104) = \log(100 + 4) \text{ and } \log(103) = \log(100 + 3) \).
Given that these approximations may not yield a simple summation directly, we check each option based more on practical understanding:
By calculating:
- \( 10^{12}= 1000000000000 \), incorrect considering scale.
- It’s clear that summing with smaller logs won't approach or equal to \( 12 \) or \( 7 \).
Ultimately utilizing rough estimations says we should incline towards \( 4 + \log 3 \), likely giving a better answer estimate re: adjacency.
Thus the closest workable option tends to be:
4 + log(3).
Please confirm back on your calculator for the precise \( log\) values as needed to cross-check completive measures!
The answer to select is \( 4 + \log 3 \).