6^log36, 81

To simplify the expression \( 6^{\log_{36}(81)} \), we can apply the change of base formula for logarithms and properties of exponents.

First, recognize that:

\[ \log_{36}(81) = \frac{\log_{6}(81)}{\log_{6}(36)} \]

Now simplify both logarithms using known values:

1. Calculate \( \log_{6}(36) \): \[ 36 = 6^2 \implies \log_{6}(36) = 2 \]

2. Calculate \( \log_{6}(81) \): Note that \( 81 = 3^4 \) and \( 3 = 6^{\log_{6}(3)} \), where \( \log_{6}(3) \) can be calculated as follows: \[ \log_{6}(3) \approx 0.5 \quad (\text{since } 6^{0.5} = \sqrt{6} \approx 2.45) \] So, \[ \log_{6}(81) = \log_{6}(3^4) = 4 \cdot \log_{6}(3) \approx 4 \cdot 0.5 = 2 \]

Now substituting back into the logarithm we have: \[ \log_{36}(81) = \frac{\log_{6}(81)}{\log_{6}(36)} = \frac{2}{2} = 1 \]

So we have: \[ 6^{\log_{36}(81)} = 6^1 = 6 \]

Therefore, the simplified result is: \[ \boxed{6} \]