1. **Problem Statement:** Find the value of $n$ in each of the following equations from question 8. 2. **Recall the exponential rules:** - $a^{m} = a^{n} \implies m = n$ if $a > 0$ and $a \neq 1$. - $(a^m)^n = a^{mn}$. - $a^{-n} = \frac{1}{a^n}$. 3. **Solve each part of question 8:** a. $(\frac{1}{3})^n = 81$ Rewrite $81$ as $3^4$ and $\frac{1}{3} = 3^{-1}$: $$3^{-n} = 3^4$$ Equate exponents: $$-n = 4 \implies n = -4$$ b. $(\frac{1}{2})^n = 8$ Rewrite $8$ as $2^3$ and $\frac{1}{2} = 2^{-1}$: $$2^{-n} = 2^3$$ Equate exponents: $$-n = 3 \implies n = -3$$ c. $(\frac{1}{2})^n = 4^{n+1}$ Rewrite $4$ as $2^2$: $$2^{-n} = (2^2)^{n+1} = 2^{2(n+1)} = 2^{2n+2}$$ Equate exponents: $$-n = 2n + 2$$ Solve for $n$: $$-n - 2n = 2 \implies -3n = 2 \implies n = -\frac{2}{3}$$ d. $(\frac{1}{2})^n = 32$ Rewrite $32$ as $2^5$ and $\frac{1}{2} = 2^{-1}$: $$2^{-n} = 2^5$$ Equate exponents: $$-n = 5 \implies n = -5$$ e. $(\frac{1}{2})^{n+1} = 2$ Rewrite $2$ as $2^1$ and $\frac{1}{2} = 2^{-1}$: $$2^{-(n+1)} = 2^1$$ Equate exponents: $$-(n+1) = 1 \implies -n -1 = 1 \implies -n = 2 \implies n = -2$$ f. $(\frac{1}{16})^n = 4$ Rewrite $16$ as $2^4$ and $4$ as $2^2$: $$(2^{-4})^n = 2^2 \implies 2^{-4n} = 2^2$$ Equate exponents: $$-4n = 2 \implies n = -\frac{1}{2}$$ --- 4. **Solve each part of question 9:** a. $3^x = 27$ Rewrite $27$ as $3^3$: $$3^x = 3^3 \implies x = 3$$ b. $4^{-x} = \frac{1}{16}$ Rewrite $16$ as $4^2$ and $\frac{1}{16} = 4^{-2}$: $$4^{-x} = 4^{-2} \implies -x = -2 \implies x = 2$$ c. $2^{-x} = 128$ Rewrite $128$ as $2^7$: $$2^{-x} = 2^7 \implies -x = 7 \implies x = -7$$ d. $2^{x+3} = 64$ Rewrite $64$ as $2^6$: $$2^{x+3} = 2^6 \implies x + 3 = 6 \implies x = 3$$ e. $3^{x+1} = \frac{1}{81}$ Rewrite $81$ as $3^4$ and $\frac{1}{81} = 3^{-4}$: $$3^{x+1} = 3^{-4} \implies x + 1 = -4 \implies x = -5$$ f. $2^{-x} = \frac{1}{256}$ Rewrite $256$ as $2^8$ and $\frac{1}{256} = 2^{-8}$: $$2^{-x} = 2^{-8} \implies -x = -8 \implies x = 8$$ --- **Final answers:** 8a. $n = -4$ 8b. $n = -3$ 8c. $n = -\frac{2}{3}$ 8d. $n = -5$ 8e. $n = -2$ 8f. $n = -\frac{1}{2}$ 9a. $x = 3$ 9b. $x = 2$ 9c. $x = -7$ 9d. $x = 3$ 9e. $x = -5$ 9f. $x = 8$