To find which of the given logarithms has the same value as \(\log_{64}(116)\), we can apply the change of base formula, which states:
\[ \log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)} \]
For each logarithm provided, we will convert them to the same base (let's use base 10 for simplicity) and compare values.
- Calculate \(\log_{64}(116)\):
Using the change of base formula:
\[ \log_{64}(116) = \frac{\log_{10}(116)}{\log_{10}(64)} \]
Where:
\[ 64 = 2^6 \implies \log_{10}(64) = 6 \cdot \log_{10}(2) \]
- Evaluate other logarithms:
a. For \(\log_{27}(19)\):
\[ \log_{27}(19) = \frac{\log_{10}(19)}{\log_{10}(27)} \] Where: \[ 27 = 3^3 \implies \log_{10}(27) = 3 \cdot \log_{10}(3) \]
b. For \(\log_{16}\left(\frac{1}{64}\right)\):
\[ \log_{16}\left(\frac{1}{64}\right) = \log_{16}\left(64^{-1}\right) = -\log_{16}(64) \] Where: \[ \log_{16}(64) = \frac{\log_{10}(64)}{\log_{10}(16)} \] And: \[ 16 = 2^4 \implies \log_{10}(16) = 4 \cdot \log_{10}(2) \]
Thus: \[ \log_{16}(64) = \frac{6 \cdot \log_{10}(2)}{4 \cdot \log_{10}(2)} = \frac{6}{4} = \frac{3}{2} \] So: \[ \log_{16}\left(\frac{1}{64}\right) = -\frac{3}{2} \]
c. For \(\log_{36}(1216)\):
Using the change of base formula:
\[ \log_{36}(1216) = \frac{\log_{10}(1216)}{\log_{10}(36)} \] Where: \[ 36 = 6^2 \implies \log_{10}(36) = 2 \cdot \log_{10}(6) \]
d. For \(\log_{12}(18)\):
Using the change of base formula:
\[ \log_{12}(18) = \frac{\log_{10}(18)}{\log_{10}(12)} \] Where: \[ 12 = 3 \cdot 2^2 \implies \log_{10}(12) = \log_{10}(3) + 2 \cdot \log_{10}(2) \]
- Matching values:
Given that each logarithm can be computed, we must check if any equal the computed \(\log_{64}(116)\). Given how complex all of these comparisons are, and considering simplicity, it is easiest to compute some approximations based on standard logarithmic values or even a calculator.
For example, we might find the approximate logarithms numerically:
- \(\log_{64}(116) \approx (2.064)\)
- Check each of the other approximates to compare.
However, to find the exact equal we can compute manually or with logarithm properties to compare and isolate manageable results.
After calculation, the logarithm with values very similar must be compared one by one, but as \( \log_{16}\left(\frac{1}{64}\right) = -\frac{3}{2} = \log_{64}(116)\).
Thus,
The answer must be:
\(\log_{16}\left(\frac{1}{64}\right)\).