1. **Problem Statement:** (a) Given $a = b^x$, $b = c^y$, and $c = a^z$, prove that $xyz = 1$. 2. **Formula and Rules:** We use the properties of exponents and logarithms. If $a = b^x$, then taking logarithm base $a$ or expressing all variables in terms of one another helps. 3. **Step-by-step Solution:** - From $a = b^x$, take logarithm base $a$: $\log_a a = \log_a b^x$ which gives $1 = x \log_a b$. - Similarly, from $b = c^y$, take logarithm base $b$: $1 = y \log_b c$. - From $c = a^z$, take logarithm base $c$: $1 = z \log_c a$. - Multiply all three equations: $$1 \times 1 \times 1 = (x \log_a b)(y \log_b c)(z \log_c a)$$ - Using change of base formula, $\log_a b \times \log_b c \times \log_c a = 1$. - Therefore, $xyz = 1$. --- 1. **Problem Statement:** (b) Find the value of $\log_2 \left(\frac{1}{128}\right)$. 2. **Formula and Rules:** Recall that $\log_b \left(\frac{1}{a}\right) = -\log_b a$ and $128 = 2^7$. 3. **Step-by-step Solution:** - $\log_2 \left(\frac{1}{128}\right) = \log_2 2^{-7} = -7$. --- 1. **Problem Statement:** (c) Given sets $A = \{x : x \in \mathbb{N}, x < 10\}$ and $B = \{x : x \in \mathbb{N}, 6 < x \leq 12\}$, find $A - B$ and $A \cap B$. 2. **Step-by-step Solution:** - $A = \{1,2,3,4,5,6,7,8,9\}$ - $B = \{7,8,9,10,11,12\}$ - $A - B$ is elements in $A$ not in $B$: $\{1,2,3,4,5,6\}$ - $A \cap B$ is common elements: $\{7,8,9\}$ --- 1. **Problem Statement:** (d) If $^{16}C_r = ^{16}C_{r+4}$, find $r$. 2. **Formula and Rules:** Recall the symmetry property of combinations: $^{n}C_r = ^{n}C_{n-r}$. 3. **Step-by-step Solution:** - Given $^{16}C_r = ^{16}C_{r+4}$ - By symmetry, $r + 4 = 16 - r$ - Solve: $2r = 12 \Rightarrow r = 6$ --- 1. **Problem Statement:** (e) Three persons enter a room with seven chairs in a line. Find the number of ways they can take their seats. 2. **Formula and Rules:** Number of ways to seat $k$ persons in $n$ chairs is $P(n,k) = \frac{n!}{(n-k)!}$. 3. **Step-by-step Solution:** - $n=7$, $k=3$ - Number of ways = $\frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5 = 210$ --- 1. **Problem Statement:** (f) A sum triples itself in 12 years at compound interest. Find the annual interest rate given $\log 3 = 0.47712$ and antilog $0.03976 = 1.096$. 2. **Formula and Rules:** Compound interest formula: $A = P(1 + r)^n$ where $A$ is amount, $P$ principal, $r$ rate per year, $n$ years. 3. **Step-by-step Solution:** - Given $A = 3P$, $n=12$ - So, $3 = (1 + r)^{12}$ - Taking logarithm base 10: $\log 3 = 12 \log (1 + r)$ - $\log (1 + r) = \frac{\log 3}{12} = \frac{0.47712}{12} = 0.03976$ - $1 + r = \text{antilog } 0.03976 = 1.096$ - Therefore, $r = 1.096 - 1 = 0.096 = 9.6\%$ per year. --- **Final answers:** (a) $xyz = 1$ (b) $\log_2 \left(\frac{1}{128}\right) = -7$ (c) $A - B = \{1,2,3,4,5,6\}$, $A \cap B = \{7,8,9\}$ (d) $r = 6$ (e) Number of ways = $210$ (f) Interest rate = $9.6\%$ per year