To solve for which logarithm has the same value as \( \log_{64}( \frac{1}{16} ) \), we can use the change of base formula for logarithms. The change of base formula states:
\[ \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \]
for any positive base \( k \).
First, let's calculate \( \log_{64}( \frac{1}{16} ) \):
Using the property of logarithms, we rewrite \( \frac{1}{16} \) as \( 16^{-1} \):
\[ \log_{64}( \frac{1}{16} ) = \log_{64}(16^{-1}) = -\log_{64}(16) \]
Now, note that \( 64 = 4^3 \) and \( 16 = 4^2 \). Therefore, we can express our logarithm using base 4:
\[ -\log_{64}(16) = -\frac{\log_4(16)}{\log_4(64)} \]
Calculating those logarithms:
- \( \log_4(16) = \log_4(4^2) = 2 \)
- \( \log_4(64) = \log_4(4^3) = 3 \)
This gives us:
\[ -\log_{64}(16) = -\frac{2}{3} = -\frac{2}{3} \]
Now we need to check which of the other given logarithms equals \( -\frac{2}{3} \):
-
For \( \log_{36}( \frac{1}{216} ) \):
\( \frac{1}{216} = 216^{-1} \)
Rewriting gives us:
\[ -\log_{36}(216) \]
We know \( 36 = 6^2 \) and \( 216 = 6^3 \). Thus:
\[ -\log_{36}(216) = -\frac{\log_6(216)}{\log_6(36)} = -\frac{3}{2} , (\text{since } \log_6(6^3) = 3 \text{ and } \log_6(6^2) = 2) \]
-
For \( \log_{1/2}( \frac{1}{8} ) \):
\( \frac{1}{8} = 8^{-1} \)
Then,
\[ -\log_{1/2}(8) \]
\( 8 = (1/2)^{-3} \), so:
\[ -\log_{1/2}(8) = -(-3) = 3 \]
-
For \( \log_{27}( \frac{1}{9} ) \):
\( \frac{1}{9} = 9^{-1} \)
This means:
\[ -\log_{27}(9) \]
Where \( 27 = 3^3 \) and \( 9 = 3^2 \), thus:
\[ -\log_{27}(9) = -\frac{\log_3(9)}{\log_3(27)} = -\frac{2}{3} \]
-
For \( \log_{16}( \frac{1}{64} ) \):
\( \frac{1}{64} = 64^{-1} \)
and thus:
\[ -\log_{16}(64) \]
Where \( 64 = 4^3 \) and \( 16 = 4^2 \), we have:
\[ -\log_{16}(64) = -\frac{\log_4(64)}{\log_4(16)} = -\frac{3}{2} \]
Thus, out of all calculated logarithms, the one with the same value as \( -\frac{2}{3} \) is:
Answer: \( \log_{27}( \frac{1}{9} ) \)