SOLUTION
1.
[tex]\begin{gathered} 5^0 \\ In\text{ indices, any entity raise to the power of zero is always 1.} \\ a^0=1 \\ \text{Therefore:} \\ 5^0=1 \end{gathered}[/tex]
3.
[tex]\begin{gathered} 24^0 \\ \text{This follows the explanation of the above question:} \\ a^0=1 \\ 24^0=1 \end{gathered}[/tex]
5.
[tex]\begin{gathered} 9^{-2} \\ To\text{ solve this, we will follow the indices law that:} \\ a^{-b}=\frac{1}{a^b} \\ \text{Therefore:} \\ 9^{-2}=\frac{1}{9^2} \\ =\frac{1}{81} \end{gathered}[/tex]
9.
[tex]\begin{gathered} 6^{-4}\cdot6^4 \\ In\text{ indices, the law that applies to the above expression:} \\ a^b\cdot a^c=a^{b+c} \\ \text{Therefore:} \\ 6^{-4}\cdot6^4 \\ =6^{-4+4} \\ =6^0 \\ =1 \end{gathered}[/tex]
7.
[tex]\begin{gathered} \frac{1}{2^{-3}} \\ In\text{ indices, this takes the form:} \\ \frac{1}{a^{-b}}=a^b \\ \text{Therefore}\colon \\ \frac{1}{2^{-3}} \\ =2^3 \\ =8 \end{gathered}[/tex]
11.
[tex]\begin{gathered} (-3\cdot2)^{-2} \\ =(-3\times2)^{-2} \\ =(-6)^{-2} \\ In\text{ indices, } \\ a^{-b}=\frac{1}{a^b} \\ \text{Therefore:} \\ =(-6)^{-2} \\ =\frac{1}{-6^2} \\ =\frac{1}{36} \end{gathered}[/tex]
