Решите пожалуйста. Мне нужно подробно.

6^(log₃₆ 81)=6^(log₆² 9²)=6^(log₆ 9)=9 5^(log₂₅ 36)=5^(log₅² 6²)=5^(log₅ 6)=6 2^(log₀₅ 3)= 2^(-1*log₂ 3)=2^(log₂ 1/3)=1/3  (0.5=1/2=2^(-1))

[latex] 6^{log_{36}81 } = 6^{log_{6^{2} }81 } =6^{ \frac{1}{2} log_{6}81 } = (6^{ log_{6}81 })^{\frac{1}{2}} = (81)^{\frac{1}{2}} = \sqrt{81} = 9 \\ 5^{log_{25}36 } =5^{log_{5^{2}}36 } =5^{\frac{1}{2} log_{5}36 } = (5^{ log_{5}36})^{\frac{1}{2}} =(36)^{\frac{1}{2}} = \sqrt{36} = 6 \\ 2^{log_{0,5}3 } = 2^{log_{2^{-1}}3 } = (2^{log_{2}3 }) ^{-1} = 3 ^{-1} = \frac{1}{3} \\ [/latex]