1. **Problem 1:** Find the smallest integral value of $x$ satisfying $$(x - 2)^{x^2 - 6x + 8} > 1.$$ 2. **Step 1:** Understand the inequality. For an expression $a^b > 1$, where $a = x-2$ and $b = x^2 - 6x + 8$, the result depends on the base $a$ and the exponent $b$. 3. **Step 2:** Factorize the exponent: $$x^2 - 6x + 8 = (x - 2)(x - 4).$$ 4. **Step 3:** Consider cases based on the base $a = x - 2$: - If $x - 2 > 1$ (i.e., $x > 3$), then since the base is greater than 1, the inequality $a^b > 1$ holds if $b > 0$. - If $x - 2 < 1$, analyze separately. 5. **Step 4:** For $x > 3$, $b > 0$ means: $$(x - 2)(x - 4) > 0 \\ ext{which implies } x < 2 \text{ or } x > 4.$$ Since $x > 3$, the valid solution is $x > 4$. 6. **Step 5:** For $x < 3$, if $x - 2 < 1$, then $b < 0$ for the inequality to hold (because a base between 0 and 1 raised to a negative power is greater than 1). 7. **Step 6:** For $b < 0$: $$(x - 2)(x - 4) < 0 \\ ext{which implies } 2 < x < 4.$$ Combined with $x < 3$, this gives $2 < x < 3$. 8. **Step 7:** Combine intervals: $$x ext{ in } (2, 3) \cup (4, \infty).$$ 9. **Step 8:** Find the smallest integral value of $x$ in these intervals: - The smallest integer in $(2, 3)$ is 3. - The smallest integer in $(4, \infty)$ is 5. Therefore, the smallest integral value satisfying the inequality is $\boxed{3}$. --- 10. **Problem 2:** Find the number of solutions to the equation: $$(2x - 3)^{2^x} = 1.$$ 11. **Step 1:** For $a^b = 1$, possible cases are: - $a = 1$ (then $1^{b} = 1$ for any $b$), - $b = 0$ (then $a^0 = 1$ for any $a \neq 0$), - $a = -1$ and $b$ is an even integer (then $(-1)^{even} = 1$). 12. **Step 2:** Analyze each case: - Case 1: $2x - 3 = 1 \Rightarrow 2x = 4 \Rightarrow x = 2$. - Case 2: $2^x = 0$ has no solution since $2^x > 0$ for all real $x$. - Case 3: $2x - 3 = -1 \Rightarrow 2x = 2 \Rightarrow x = 1$. Check if $b = 2^x$ is an even integer at $x=1$: $2^1 = 2$, which is even integer, so $x=1$ is a solution. 13. **Step 3:** Check if there are other solutions by solving: $$(2x - 3)^{2^x} = 1 \Rightarrow 2x - 3 = 2^{-x}$$ This is a transcendental equation; graphically, the number of solutions equals the number of intersections of: - $y = 2x - 3$ (a line), - $y = 2^{-x}$ (an exponential decay). 14. **Step 4:** From the graph shape and analysis: - The line crosses $y=1$ at $x=2$ (solution 1). - The line crosses $y=-1$ at $x=1$ (solution 2). - The graphs intersect at exactly two points. 15. **Final answer:** There are exactly $\boxed{2}$ solutions to the equation.