Question
Evaluate log(1011⋅105)
Answers
GPT-4o mini
To evaluate \( \log(1011 \cdot 105) \), we can use the property of logarithms that states:
\[
\log(a \cdot b) = \log a + \log b
\]
Applying this property, we get:
\[
\log(1011 \cdot 105) = \log(1011) + \log(105)
\]
Next, we can compute \( \log(1011) \) and \( \log(105) \) using a calculator (or logarithm tables) or approximate these values using known logarithms.
Using approximate values:
\[
\log(1011) \approx 3.0043 \quad \text{(since 1011 is slightly more than 1000)}
\]
\[
\log(105) \approx 2.0212 \quad \text{(since \( 10^2 = 100 \) and \( 10^3 = 1000 \), and \( 105 \) is slightly above \( 100 \))}
\]
Now we can add these two results together:
\[
\log(1011 \cdot 105) \approx 3.0043 + 2.0212 = 5.0255
\]
Therefore, the final evaluation gives us:
\[
\log(1011 \cdot 105) \approx 5.0255
\]
So, the answer is approximately:
\[
\boxed{5.0255}
\]
\[
\log(a \cdot b) = \log a + \log b
\]
Applying this property, we get:
\[
\log(1011 \cdot 105) = \log(1011) + \log(105)
\]
Next, we can compute \( \log(1011) \) and \( \log(105) \) using a calculator (or logarithm tables) or approximate these values using known logarithms.
Using approximate values:
\[
\log(1011) \approx 3.0043 \quad \text{(since 1011 is slightly more than 1000)}
\]
\[
\log(105) \approx 2.0212 \quad \text{(since \( 10^2 = 100 \) and \( 10^3 = 1000 \), and \( 105 \) is slightly above \( 100 \))}
\]
Now we can add these two results together:
\[
\log(1011 \cdot 105) \approx 3.0043 + 2.0212 = 5.0255
\]
Therefore, the final evaluation gives us:
\[
\log(1011 \cdot 105) \approx 5.0255
\]
So, the answer is approximately:
\[
\boxed{5.0255}
\]