\( \begin{array}{l} 8^{-2}=\frac{1}{64}\\ \text { h) }\\ 0=\log _{6} 1\\ \begin{array}{ll} \log _{x} x^{4}=4 \\ m^{n}=p & \log _{6}\left(\frac{1}{36}\right)=-2 \\ 3^{0}=1 & 2=\log _{1} x \\ 10^{-3}=0,001 & \\ \frac{1}{25}=5^{-2} & \\ \left(\frac{1}{2}\right)^{-2}=4 & \end{array} \end{array} \)

Let's analyze the given equations and statements step by step. 1. **First Statement:** \[ 8^{-2} = \frac{1}{64} \] This is correct. We can verify it: \[ 8^{-2} = \frac{1}{8^2} = \frac{1}{64} \] 2. **Second Statement:** \[ 0 = \log_{6} 1 \] This is also correct. The logarithm of 1 in any base is 0: \[ \log_{b} 1 = 0 \quad \text{for any } b > 0 \] 3. **Third Statement:** \[ \log_{x} x^{4} = 4 \] This is true. By the properties of logarithms: \[ \log_{b} b^{k} = k \quad \text{so here } k = 4 \] 4. **Fourth Statement:** \[ m^{n} = p \] This is a general expression and does not require verification without specific values. 5. **Fifth Statement:** \[ \log_{6}\left(\frac{1}{36}\right) = -2 \] We can verify this: \[ \frac{1}{36} = 6^{-2} \implies \log_{6}(6^{-2}) = -2 \] 6. **Sixth Statement:** \[ 3^{0} = 1 \] This is correct. Any non-zero number raised to the power of 0 is 1. 7. **Seventh Statement:** \[ 2 = \log_{1} x \] This statement is incorrect. The logarithm base 1 is undefined because you cannot raise 1 to any power to get a number other than 1. 8. **Eighth Statement:** \[ 10^{-3} = 0.001 \] This is correct. We can verify it: \[ 10^{-3} = \frac{1}{10^3} = \frac{1}{1000} = 0.001 \] 9. **Ninth Statement:** \[ \frac{1}{25} = 5^{-2} \] This is correct. We can verify it: \[ \frac{1}{25} = \frac{1}{5^2} = 5^{-2} \] 10. **Tenth Statement:** \[ \left(\frac{1}{2}\right)^{-2} = 4 \] This is correct. We can verify it: \[ \left(\frac{1}{2}\right)^{-2} = \left(2\right)^{2} = 4 \] ### Summary of Validity: - Valid: 1, 2, 3, 5, 6, 8, 9, 10 - Invalid: 7 If you have any specific questions or need further assistance with any of these statements, feel free to ask!