Solution:
The equation is given below as
[tex]\begin{gathered} \log_2x=-5 \\ \log_ab=y \\ b=a^y \end{gathered}[/tex]
By applying the law above, we will have
[tex]\begin{gathered} \log_2x=-5 \\ x=2^{-5} \\ x=\frac{1}{2^5} \\ x=\frac{1}{32} \end{gathered}[/tex]
Hence,
The final answer is
[tex]\Rightarrow x=\frac{1}{32}[/tex]
Part B:
The expression is given below as
[tex]\begin{gathered} \log_4(\frac{1}{64}) \\ \log_4(64^{-1}) \\ \log_4(4^{-3}) \\ -3\log_44 \\ =-3\times1 \\ =-3 \end{gathered}[/tex]
Hence,
The final answer is
[tex]\Rightarrow\log_4(\frac{1}{64})=-3[/tex]
