a) \( 6^{\log _{6} 0,125} \) 8) \( 4^{\log _{4} 10-3} \)
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To solve \( 6^{\log_{6} 0.125} \), we can use the property of logarithms which states that \( a^{\log_{a} b} = b \). So, \( 6^{\log_{6} 0.125} = 0.125 \). For the second expression, \( 4^{\log_{4} 10-3} \), we first rewrite it using logarithm properties. This becomes \( 4^{\log_{4} 10 - \log_{4} 4^3} = 4^{\log_{4} \left( \frac{10}{64} \right)} \). Using the same property, we can simplify this to \( \frac{10}{64} = \frac{5}{32} \). Therefore, \( 4^{\log_{4} 10-3} = \frac{5}{32} \). So the final results are: a) \( 0.125 \) b) \( \frac{5}{32} \)