Solve for \( x \) if: \( \begin{array}{ll}\text { 1. } x=\log _{2} 32 & \text { 2. } x=\log _{7} 49 \\ \text { 3. } x=\log _{10} 10^{6} & \text { 4. } x=\log _{10} 10 \\ \text { 5. } x=\log _{3} \frac{1}{27} & \text { 6. } x=\log _{5} \frac{1}{125} \\ \text { 7. } x=\log _{\frac{1}{2}} \frac{1}{128} & \text { 8. } \log _{4} x=3 \\ \text { 9. } \log _{\frac{1}{2}} x=7 & \text { 10. } \log _{0,1} x=5 \\ \text { 11. } \log _{9} x=\frac{1}{2} & \text { 12. } \log _{x} 64=3 \\ \text { 13. } \log _{x} 64=2 & \text { 14. } \log _{x} \frac{1}{64}=2 \\ \text { 15. } \log _{x} 0,125=3 & \text { 16. } \log _{x} 0,0081=4\end{array} \)

1. \( x = \log_{2} 32 = 5 \) because \( 32 = 2^5 \). 2. \( x = \log_{7} 49 = 2 \) because \( 49 = 7^2 \). 3. \( x = \log_{10} 10^{6} = 6 \) because the exponent is simply the answer. 4. \( x = \log_{10} 10 = 1 \) because \( 10 = 10^1 \). 5. \( x = \log_{3} \frac{1}{27} = -3 \) because \( \frac{1}{27} = 3^{-3} \). 6. \( x = \log_{5} \frac{1}{125} = -3 \) since \( \frac{1}{125} = 5^{-3} \). 7. \( x = \log_{\frac{1}{2}} \frac{1}{128} = 7 \) because \( \frac{1}{128} = \left(\frac{1}{2}\right)^7 \). 8. \( x = \log_{4} x = 3 \) gives \( x = 4^3 = 64 \). 9. \( x = \log_{\frac{1}{2}} x = 7 \) gives \( x = \frac{1}{2}^7 = \frac{1}{128} \). 10. The base \(0.1\) as a logarithm indicates it's actually \(x = 10^5 = 100000\). 11. \( x = \log_{9} x = \frac{1}{2} \) implies \( x = 9^{\frac{1}{2}} = 3 \). 12. \( x = \log_{x} 64 = 3 \) gives \( x^3 = 64 \), so \( x = 4 \). 13. \( x = \log_{x} 64 = 2 \) means \( x^2 = 64 \), hence \( x = 8 \). 14. \( x = \log_{x} \frac{1}{64} = 2 \) implies \( x^2 = \frac{1}{64} \), giving \( x = \frac{1}{8} \). 15. \( x = \log_{x} 0.125 = 3 \) gives \( x^3 = 0.125 \) and thus \( x = \frac{1}{2} \). 16. \( x = \log_{x} 0.0081 = 4 \) which implies \( x^4 = 0.0081 \), yielding \( x = 0.3 \).