Question
Rewrite the expression 2log636−log864
2
log
6
36
−
log
8
64
using the properties of logarithms.(1 point)
Responses
0
0
log−2(36264)
log
−
2
(
36
2
64
)
logarithm subscript negative 2 baseline left parenthesis Start Fraction 36 squared over 64 End Fraction right parenthesis
4
4
2
2
log
6
36
−
log
8
64
using the properties of logarithms.(1 point)
Responses
0
0
log−2(36264)
log
−
2
(
36
2
64
)
logarithm subscript negative 2 baseline left parenthesis Start Fraction 36 squared over 64 End Fraction right parenthesis
4
4
2
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To rewrite the expression \(2 \log_6 36 - \log_8 64\) using the properties of logarithms, we can apply the following properties:
1. The power rule: \(a \log_b x = \log_b (x^a)\)
2. The change of base rule: \(\log_b x = \frac{\log_k x}{\log_k b}\) (for this problem, we can keep the logarithms in their original bases if applicable)
Let's proceed step by step:
1. **Apply the power rule to \(2 \log_6 36\)**:
\[
2 \log_6 36 = \log_6 (36^2)
\]
2. **Calculate \(36^2\)**:
\[
36^2 = 1296
\]
Thus,
\[
2 \log_6 36 = \log_6 1296
\]
3. **Now consider \(\log_8 64\)**. Since \(64 = 8^2\), we can use the property of logarithms:
\[
\log_8 64 = \log_8 (8^2) = 2
\]
4. **Combine the two parts together**:
We now substitute back into the original expression:
\[
2 \log_6 36 - \log_8 64 = \log_6 1296 - 2
\]
5. **Express the subtraction in terms of logarithms**:
We can rewrite \(2\) as \(\log_6 (6^2)\), since \(2 = \log_6(36)\):
\[
2 = \log_6 36
\]
Therefore:
\[
\log_6 1296 - \log_6 36 = \log_6 \left(\frac{1296}{36}\right)
\]
6. **Calculate \(\frac{1296}{36}\)**:
\[
\frac{1296}{36} = 36
\]
Hence,
\[
\log_6 \left(\frac{1296}{36}\right) = \log_6 36
\]
This gives us the final expression as:
\[
\log_6 36
\]
So the rewritten expression can be presented as:
\[
\log_6 \left(\frac{36^2}{64}\right)
\]
or simply,
\[
\log_6 36
\]
Thus, the responses:
- **Option 1**: \(\log_{-2}\left(\frac{36^2}{64}\right)\)
- **Option 4**: \(4\) (equivalent, if we evaluate \(\log_6 36\))
Based on the responses you provided:
- The expression corresponds to the first interpretation, but note that \(4\) is incorrect in evaluation after steps since that yields finite simple terms.
Please clarify if you need a particular option in the set responses you mentioned!
1. The power rule: \(a \log_b x = \log_b (x^a)\)
2. The change of base rule: \(\log_b x = \frac{\log_k x}{\log_k b}\) (for this problem, we can keep the logarithms in their original bases if applicable)
Let's proceed step by step:
1. **Apply the power rule to \(2 \log_6 36\)**:
\[
2 \log_6 36 = \log_6 (36^2)
\]
2. **Calculate \(36^2\)**:
\[
36^2 = 1296
\]
Thus,
\[
2 \log_6 36 = \log_6 1296
\]
3. **Now consider \(\log_8 64\)**. Since \(64 = 8^2\), we can use the property of logarithms:
\[
\log_8 64 = \log_8 (8^2) = 2
\]
4. **Combine the two parts together**:
We now substitute back into the original expression:
\[
2 \log_6 36 - \log_8 64 = \log_6 1296 - 2
\]
5. **Express the subtraction in terms of logarithms**:
We can rewrite \(2\) as \(\log_6 (6^2)\), since \(2 = \log_6(36)\):
\[
2 = \log_6 36
\]
Therefore:
\[
\log_6 1296 - \log_6 36 = \log_6 \left(\frac{1296}{36}\right)
\]
6. **Calculate \(\frac{1296}{36}\)**:
\[
\frac{1296}{36} = 36
\]
Hence,
\[
\log_6 \left(\frac{1296}{36}\right) = \log_6 36
\]
This gives us the final expression as:
\[
\log_6 36
\]
So the rewritten expression can be presented as:
\[
\log_6 \left(\frac{36^2}{64}\right)
\]
or simply,
\[
\log_6 36
\]
Thus, the responses:
- **Option 1**: \(\log_{-2}\left(\frac{36^2}{64}\right)\)
- **Option 4**: \(4\) (equivalent, if we evaluate \(\log_6 36\))
Based on the responses you provided:
- The expression corresponds to the first interpretation, but note that \(4\) is incorrect in evaluation after steps since that yields finite simple terms.
Please clarify if you need a particular option in the set responses you mentioned!
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