Asked by 54t4n45
Evaluate log(104⋅8)
.(1 point)
Responses
log(104⋅8)=4+log8
logarithm left parenthesis 10 superscript 4 baseline times 8 right parenthesis equals 4 plus logarithm 8
log(104⋅8)=log(4⋅8)
logarithm left parenthesis 10 superscript 4 baseline times 8 right parenthesis equals logarithm left parenthesis 4 times 8 right parenthesis
log(104⋅8)=log4+8
logarithm left parenthesis 10 superscript 4 baseline times 8 right parenthesis equals logarithm 4 plus 8
log(104⋅8)=log(4+8)
.(1 point)
Responses
log(104⋅8)=4+log8
logarithm left parenthesis 10 superscript 4 baseline times 8 right parenthesis equals 4 plus logarithm 8
log(104⋅8)=log(4⋅8)
logarithm left parenthesis 10 superscript 4 baseline times 8 right parenthesis equals logarithm left parenthesis 4 times 8 right parenthesis
log(104⋅8)=log4+8
logarithm left parenthesis 10 superscript 4 baseline times 8 right parenthesis equals logarithm 4 plus 8
log(104⋅8)=log(4+8)
Answers
Answered by
GPT-4o mini
To evaluate \( \log(104 \cdot 8) \), we can use the properties of logarithms.
The property of logarithms we will use here is:
\[
\log(a \cdot b) = \log a + \log b
\]
Applying this property, we have:
\[
\log(104 \cdot 8) = \log(104) + \log(8)
\]
Next, we can break down \( 104 \) further:
\[
104 = 10^2 \cdot 10^1 + 4 = 100 + 4 = 104
\]
Using the property again:
\[
\log(104) = \log(100 + 4) \text{ which we cannot simplify directly, so we stick to the original.}
\]
Since we know:
\[
\log(100) = 2 \quad \text{and we approximate } \log(4) \approx 0.6021 \text{ and } \log(8) \approx 0.903
\]
We can recognize that \( \log(8) = \log(2^3) = 3\log(2) \).
In simpler terms, without specific computation, we recognize that:
\[
\log(104 \cdot 8) = \log(104) + \log(8) \quad \Rightarrow \quad \text{which is not one of the options.}
\]
Now, focusing on the closest options presented:
1. \( \log(104 \cdot 8) = 4 + \log(8) \) - This reconstruction is incorrect based on simplifications.
2. \( \log(104 \cdot 8) = \log(4 \cdot 8) \) - This reconstruction is incorrect.
3. \( \log(104 \cdot 8) = \log4 + 8 \) - This reconstruction is incorrect.
4. \( \log(104 \cdot 8) = \log(4 + 8) \) - This reconstruction is incorrect.
Thus, the correct expression should indeed be:
\[
\log(104 \cdot 8) = \log(104) + \log(8)
\]
However, referring back to the solutions proposed: the answer is closest to:
**\( \log(104 \cdot 8) = 4 + \log(8) \)**, as it separates out \( 100 \) from \( 4 \cdot 8 \) rather than attempting to sum directly.
Thus:
**\( \log(104 \cdot 8) = 4 + \log(8) \)** is the most accurate from the options provided.
The property of logarithms we will use here is:
\[
\log(a \cdot b) = \log a + \log b
\]
Applying this property, we have:
\[
\log(104 \cdot 8) = \log(104) + \log(8)
\]
Next, we can break down \( 104 \) further:
\[
104 = 10^2 \cdot 10^1 + 4 = 100 + 4 = 104
\]
Using the property again:
\[
\log(104) = \log(100 + 4) \text{ which we cannot simplify directly, so we stick to the original.}
\]
Since we know:
\[
\log(100) = 2 \quad \text{and we approximate } \log(4) \approx 0.6021 \text{ and } \log(8) \approx 0.903
\]
We can recognize that \( \log(8) = \log(2^3) = 3\log(2) \).
In simpler terms, without specific computation, we recognize that:
\[
\log(104 \cdot 8) = \log(104) + \log(8) \quad \Rightarrow \quad \text{which is not one of the options.}
\]
Now, focusing on the closest options presented:
1. \( \log(104 \cdot 8) = 4 + \log(8) \) - This reconstruction is incorrect based on simplifications.
2. \( \log(104 \cdot 8) = \log(4 \cdot 8) \) - This reconstruction is incorrect.
3. \( \log(104 \cdot 8) = \log4 + 8 \) - This reconstruction is incorrect.
4. \( \log(104 \cdot 8) = \log(4 + 8) \) - This reconstruction is incorrect.
Thus, the correct expression should indeed be:
\[
\log(104 \cdot 8) = \log(104) + \log(8)
\]
However, referring back to the solutions proposed: the answer is closest to:
**\( \log(104 \cdot 8) = 4 + \log(8) \)**, as it separates out \( 100 \) from \( 4 \cdot 8 \) rather than attempting to sum directly.
Thus:
**\( \log(104 \cdot 8) = 4 + \log(8) \)** is the most accurate from the options provided.
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