The Solution.
[tex]Let\log _{\frac{1}{9}}(\frac{1}{9})=x[/tex]Writing the above equation in index form, we have
[tex]\begin{gathered} (\frac{1}{9})^1=(\frac{1}{9})^x \\ Then\text{ it follows that} \\ 1=x \\ x=1 \end{gathered}[/tex][tex]\begin{gathered} \text{Let }\log _749=x \\ So\text{ we have} \\ 49=7^x \\ \text{Making the base of both sides equal, we have} \\ 7^2=7^x \\ x=2 \end{gathered}[/tex][tex]\begin{gathered} \text{Let }\log _{\frac{1}{4}}16=x \\ \\ 16=(\frac{1}{4})^x \\ 16=4^{-1\times x} \\ 4^2=4^{-x} \\ -x=2 \\ x=-2 \end{gathered}[/tex][tex]\begin{gathered} \text{Let }\log _{125}5=x \\ \text{cross multiplying, we have} \\ 5=125^x \\ 5^1=5^{3x} \\ 3x=1 \\ \text{Dividing both sides by 3, we get} \\ x=\frac{1}{3} \end{gathered}[/tex][tex]\begin{gathered} \text{ Let }\log _8(\frac{1}{8})=x \\ \\ \frac{1}{8}=8^x \\ \\ 8^{-1}=8^x \\ -1=x \\ x=-1 \end{gathered}[/tex][tex]\begin{gathered} \text{ Let }\log _9(1)=x \\ 1=9^x \\ 9^0=9^x \\ x=0 \end{gathered}[/tex][tex]\begin{gathered} \text{ Let }\log _{\frac{1}{9}}(-1)=x \\ \\ -1=(\frac{1}{9})^x \\ \\ No\text{ solution because it has no real value.} \end{gathered}[/tex]