1. **Problem (ক):** Given $a^x = b$, $b^y = c$, and $c^z = a$, show that $xyz = 1$. 2. **Step 1:** Express $b$ and $c$ in terms of $a$: $$b = a^x$$ $$c = b^y = (a^x)^y = a^{xy}$$ 3. **Step 2:** Use $c^z = a$: $$c^z = (a^{xy})^z = a^{xyz} = a$$ 4. **Step 3:** Since bases are equal and $a > 0$, equate exponents: $$xyz = 1$$ --- 1. **Problem (খ):** Given $P^w = Q^v = R^u$ where $P = a^x$, $Q = a^y$, $R = a^z$, prove that $x/u = y/v = z/w$. 2. **Step 1:** Write each expression in terms of $a$: $$P^w = (a^x)^w = a^{xw}$$ $$Q^v = (a^y)^v = a^{yv}$$ $$R^u = (a^z)^u = a^{zu}$$ 3. **Step 2:** Since $P^w = Q^v = R^u$, their exponents must be equal: $$xw = yv = zu$$ 4. **Step 3:** Divide each equality to isolate ratios: $$\frac{x}{u} = \frac{y}{v} = \frac{z}{w}$$ --- 1. **Problem (গ):** Given $x^2 + y^2 = 7xy$, show that $$\log \frac{x + y}{3} = \frac{1}{2} (\log x + \log y)$$ 2. **Step 1:** Start with the given equation: $$x^2 + y^2 = 7xy$$ 3. **Step 2:** Divide both sides by $xy$: $$\frac{x^2}{xy} + \frac{y^2}{xy} = \frac{7xy}{xy}$$ $$\frac{x}{y} + \frac{y}{x} = 7$$ 4. **Step 3:** Let $t = \frac{x}{y}$, then: $$t + \frac{1}{t} = 7$$ Multiply both sides by $t$: $$t^2 + 1 = 7t$$ Rearranged: $$t^2 - 7t + 1 = 0$$ 5. **Step 4:** The quadratic in $t$ has roots $t_1$ and $t_2$ such that $t_1 t_2 = 1$. 6. **Step 5:** Now, consider the expression to prove: $$\log \frac{x + y}{3} = \frac{1}{2} (\log x + \log y)$$ Rewrite right side: $$\frac{1}{2} (\log x + \log y) = \frac{1}{2} \log (xy) = \log \sqrt{xy}$$ 7. **Step 6:** So, we need to show: $$\frac{x + y}{3} = \sqrt{xy}$$ 8. **Step 7:** Square both sides: $$\left(\frac{x + y}{3}\right)^2 = xy$$ $$\frac{(x + y)^2}{9} = xy$$ Multiply both sides by 9: $$(x + y)^2 = 9xy$$ 9. **Step 8:** Expand left side: $$x^2 + 2xy + y^2 = 9xy$$ 10. **Step 9:** Rearrange: $$x^2 + y^2 = 7xy$$ Which matches the given condition, so the equality holds. --- **Final answers:** (ক) $xyz = 1$ (খ) $\frac{x}{u} = \frac{y}{v} = \frac{z}{w}$ (গ) $\log \frac{x + y}{3} = \frac{1}{2} (\log x + \log y)$