Question
Which logarithm has the same value as log64 (1/16)
Log16 (1/64)
Log1/2 (1/8)
Log27 (1/9)
Log36 (1/216)
Answers
GPT-4o mini
To determine which logarithm has the same value as \( \log_{64} \left(\frac{1}{16}\right) \), we can first simplify \( \log_{64} \left(\frac{1}{16}\right) \).
Using the change of base formula:
\[
\log_{64} \left(\frac{1}{16}\right) = \frac{\log \left(\frac{1}{16}\right)}{\log(64)}
\]
Next, we express both 64 and \( \frac{1}{16} \) as powers of 2:
\[
64 = 2^6 \quad \text{and} \quad \frac{1}{16} = 16^{-1} = (2^4)^{-1} = 2^{-4}
\]
Now we can calculate:
\[
\log_{64} \left(\frac{1}{16}\right) = \log_{2^6} (2^{-4})
\]
Using the change of base formula for logarithms:
\[
\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)}
\]
We can simplify it further:
\[
\log_{2^6}(2^{-4}) = \frac{-4}{6} = -\frac{2}{3}
\]
Now we need to check each option to see which one also equals \(-\frac{2}{3}\).
1. **For** \( \log_{16} \left(\frac{1}{64}\right) \):
\[
16 = 2^4, \quad \text{and} \quad \frac{1}{64} = 64^{-1} = (2^6)^{-1} = 2^{-6}
\]
\[
\log_{16} \left(\frac{1}{64}\right) = \log_{2^4} (2^{-6}) = \frac{-6}{4} = -\frac{3}{2}
\]
2. **For** \( \log_{1/2} \left( \frac{1}{8} \right) \):
\[
\frac{1}{8} = 8^{-1} = (2^3)^{-1} = 2^{-3}
\]
\[
\log_{1/2} \left(\frac{1}{8}\right) = \log_{2^{-1}} (2^{-3}) = \frac{-3}{-1} = 3
\]
3. **For** \( \log_{27} \left( \frac{1}{9} \right) \):
\[
27 = 3^3, \quad \text{and} \quad \frac{1}{9} = 9^{-1} = (3^2)^{-1} = 3^{-2}
\]
\[
\log_{27} \left(\frac{1}{9}\right) = \log_{3^3}(3^{-2}) = \frac{-2}{3}
\]
4. **For** \( \log_{36} \left( \frac{1}{216} \right) \):
\[
36 = 6^2, \quad \text{and} \quad \frac{1}{216} = 216^{-1} = (6^3)^{-1} = 6^{-3}
\]
\[
\log_{36} \left(\frac{1}{216}\right) = \log_{6^2} (6^{-3}) = \frac{-3}{2}
\]
Now, comparing the values:
- \( \log_{16} \left(\frac{1}{64}\right) = -\frac{3}{2} \)
- \( \log_{1/2} \left(\frac{1}{8}\right) = 3 \)
- \( \log_{27} \left(\frac{1}{9}\right) = -\frac{2}{3} \) **(Matches)**
- \( \log_{36} \left(\frac{1}{216}\right) = -\frac{3}{2} \)
The logarithm that has the same value as \( \log_{64} \left(\frac{1}{16}\right) \) is:
\[
\boxed{\log_{27} \left( \frac{1}{9} \right)}
\]
Using the change of base formula:
\[
\log_{64} \left(\frac{1}{16}\right) = \frac{\log \left(\frac{1}{16}\right)}{\log(64)}
\]
Next, we express both 64 and \( \frac{1}{16} \) as powers of 2:
\[
64 = 2^6 \quad \text{and} \quad \frac{1}{16} = 16^{-1} = (2^4)^{-1} = 2^{-4}
\]
Now we can calculate:
\[
\log_{64} \left(\frac{1}{16}\right) = \log_{2^6} (2^{-4})
\]
Using the change of base formula for logarithms:
\[
\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)}
\]
We can simplify it further:
\[
\log_{2^6}(2^{-4}) = \frac{-4}{6} = -\frac{2}{3}
\]
Now we need to check each option to see which one also equals \(-\frac{2}{3}\).
1. **For** \( \log_{16} \left(\frac{1}{64}\right) \):
\[
16 = 2^4, \quad \text{and} \quad \frac{1}{64} = 64^{-1} = (2^6)^{-1} = 2^{-6}
\]
\[
\log_{16} \left(\frac{1}{64}\right) = \log_{2^4} (2^{-6}) = \frac{-6}{4} = -\frac{3}{2}
\]
2. **For** \( \log_{1/2} \left( \frac{1}{8} \right) \):
\[
\frac{1}{8} = 8^{-1} = (2^3)^{-1} = 2^{-3}
\]
\[
\log_{1/2} \left(\frac{1}{8}\right) = \log_{2^{-1}} (2^{-3}) = \frac{-3}{-1} = 3
\]
3. **For** \( \log_{27} \left( \frac{1}{9} \right) \):
\[
27 = 3^3, \quad \text{and} \quad \frac{1}{9} = 9^{-1} = (3^2)^{-1} = 3^{-2}
\]
\[
\log_{27} \left(\frac{1}{9}\right) = \log_{3^3}(3^{-2}) = \frac{-2}{3}
\]
4. **For** \( \log_{36} \left( \frac{1}{216} \right) \):
\[
36 = 6^2, \quad \text{and} \quad \frac{1}{216} = 216^{-1} = (6^3)^{-1} = 6^{-3}
\]
\[
\log_{36} \left(\frac{1}{216}\right) = \log_{6^2} (6^{-3}) = \frac{-3}{2}
\]
Now, comparing the values:
- \( \log_{16} \left(\frac{1}{64}\right) = -\frac{3}{2} \)
- \( \log_{1/2} \left(\frac{1}{8}\right) = 3 \)
- \( \log_{27} \left(\frac{1}{9}\right) = -\frac{2}{3} \) **(Matches)**
- \( \log_{36} \left(\frac{1}{216}\right) = -\frac{3}{2} \)
The logarithm that has the same value as \( \log_{64} \left(\frac{1}{16}\right) \) is:
\[
\boxed{\log_{27} \left( \frac{1}{9} \right)}
\]