Solve for \( x \) if: 1. \( x=\log _{2} 32 \) 3. \( x=\log _{10} 10^{6} \) 5. \( x=\log _{3} \frac{1}{27} \) 7. \( x=\log _{\frac{1}{2}} \frac{1}{128} \) 9. \( \log _{\frac{1}{2}} x=7 \) 11. \( \log _{9} x=\frac{1}{2} \) 13. \( \log _{x} 64=2 \) 15. \( \log _{x} 0,125=3 \)

Let's solve each logarithmic equation step by step: 1. \( x = \log_{2} 32 \) - Since \( 32 = 2^5 \), we have \( x = 5 \). 2. \( x = \log_{10} 10^{6} \) - By the properties of logarithms, \( x = 6 \). 3. \( x = \log_{3} \frac{1}{27} \) - \( \frac{1}{27} = 3^{-3} \), so \( x = -3 \). 4. \( x = \log_{\frac{1}{2}} \frac{1}{128} \) - Since \( \frac{1}{128} = \left( \frac{1}{2} \right)^7 \), we get \( x = 7 \). 5. \( \log_{\frac{1}{2}} x = 7 \) - This means \( x = \left( \frac{1}{2} \right)^{-7} = 2^7 = 128 \). 6. \( \log_{9} x = \frac{1}{2} \) - This indicates \( x = 9^{\frac{1}{2}} = 3 \). 7. \( \log_{x} 64 = 2 \) - This means \( 64 = x^2 \). Since \( 64 = 8^2 \) or \( 64 = 2^6 \), \( x = 8 \) or \( x = -8 \) (but \( x \) must be positive, so \( x = 8 \)). 8. \( \log_{x} 0.125 = 3 \) - Here, \( 0.125 = \frac{1}{8} = 8^{-1} \), leading us to \( x^3 = 0.125 \) which gives \( x = 8^{-1/3} = \frac{1}{2} \). In summary, the solutions are: 1. \( x = 5 \) 2. \( x = 6 \) 3. \( x = -3 \) 4. \( x = 7 \) 5. \( x = 128 \) 6. \( x = 3 \) 7. \( x = 8 \) 8. \( x = \frac{1}{2} \)