1. **Problem statement:** (a) Write each number 64, 1/16, and 2 in the form $4^r$ where $r \in \mathbb{Q}$. (b)(i) Place the numbers 64, 1/16, $\sqrt{3}$, $\sqrt{2}$, 12, and 1 in the correct regions of the Venn diagram with sets: - $A$: all integer multiples of 4 ($4m$, $m \in \mathbb{Z}$) - $B$: all numbers of the form $4^r$, $r \in \mathbb{Q}$ - $D$: irrational numbers ($\mathbb{R} \setminus \mathbb{Q}$) (b)(ii) Find one real number not in $A$, $B$, or $D$ (not using the numbers from part (b)(i)). (c) For real $k$, solve $4^k = 9$ and express $k$ in the form $\frac{a}{\log_3 b}$ where $a,b \in \mathbb{N}$. --- 2. **Formulas and rules:** - Any positive number can be expressed as $4^r$ if $r = \log_4$(number). - $4 = 2^2$, so $4^r = (2^2)^r = 2^{2r}$. - Rational exponents $r$ can be fractions. - $A \cap D = \emptyset$ since multiples of 4 are integers (rational), so no integer multiple of 4 is irrational. --- 3. **Part (a) - Express numbers as $4^r$:** - $64 = 4^r$ means $64 = (2^2)^r = 2^{2r}$. Since $64 = 2^6$, we have $2^6 = 2^{2r} \Rightarrow 6 = 2r \Rightarrow r = 3$. - $\frac{1}{16} = 4^r$ means $\frac{1}{16} = 2^{-4} = 2^{2r}$. So $2^{-4} = 2^{2r} \Rightarrow -4 = 2r \Rightarrow r = -2$. - $2 = 4^r = 2^{2r}$. Since $2 = 2^1$, then $1 = 2r \Rightarrow r = \frac{1}{2}$. Thus: $$ 64 = 4^3, \quad \frac{1}{16} = 4^{-2}, \quad 2 = 4^{\frac{1}{2}} $$ --- 4. **Part (b)(i) - Place numbers in Venn diagram:** - $64 = 4^3$ is in $A$ (multiple of 4) and in $B$ (of form $4^r$). - $\frac{1}{16} = 4^{-2}$ is in $B$ only (not integer multiple of 4). - $\sqrt{3}$ is irrational, so in $D$ only. - $\sqrt{2}$ is irrational, so in $D$ only. - $12$ is an integer but not multiple of 4, so in $\mathbb{Q}$ but not in $A$ or $B$. - $1 = 4^0$ is in $B$ only (not multiple of 4). Summary: - $64$ in $A \cap B$ - $\frac{1}{16}$ and $1$ in $B$ only - $\sqrt{3}$ and $\sqrt{2}$ in $D$ - $12$ in none of $A$, $B$, or $D$ --- 5. **Part (b)(ii) - Number not in $A$, $B$, or $D$:** - $12$ is one such number (already used). - Another example: $5$ (integer, not multiple of 4, not of form $4^r$, and rational, so not in $D$). --- 6. **Part (c) - Solve $4^k = 9$:** Rewrite $4 = 2^2$ and $9 = 3^2$: $$ (2^2)^k = 3^2 \Rightarrow 2^{2k} = 3^2 $$ Take logarithm base 3: $$ \log_3(2^{2k}) = \log_3(3^2) \Rightarrow 2k \log_3 2 = 2 $$ Divide both sides by $2 \log_3 2$: $$ k = \frac{2}{2 \log_3 2} = \frac{1}{\log_3 2} $$ Rewrite $k$ as: $$ k = \frac{1}{\log_3 2} $$ Here $a=1$, $b=2$ are natural numbers. --- **Final answers:** (a) $$ 64 = 4^3, \quad \frac{1}{16} = 4^{-2}, \quad 2 = 4^{\frac{1}{2}} $$ (b)(i) Placement: - $64 \in A \cap B$ - $\frac{1}{16}, 1 \in B$ - $\sqrt{3}, \sqrt{2} \in D$ - $12$ outside $A$, $B$, $D$ (b)(ii) Example number not in $A$, $B$, or $D$: $5$ (c) $$ k = \frac{1}{\log_3 2} $$