1. **Problem 1: Find the smallest integral value of $x$ satisfying $$(x - 2)^{x^2 - 6x + 8} > 1.$$ 2. **Step 1: Understand the inequality** We want to find $x$ such that the expression is greater than 1. Since the base and exponent depend on $x$, we analyze cases based on the base $(x-2)$. 3. **Step 2: Case when $x - 2 > 1$ (i.e., $x > 3$)** If the base is greater than 1, then for the power to be greater than 1, the exponent must be positive: $$x^2 - 6x + 8 > 0.$$ Factor the quadratic: $$(x - 2)(x - 4) > 0.$$ This inequality holds when $x < 2$ or $x > 4$. Since $x > 3$ from the base condition, combining gives $x > 4$. 4. **Step 3: Case when $0 < x - 2 < 1$ (i.e., $2 < x < 3$)** If the base is between 0 and 1, the inequality $$(x - 2)^{x^2 - 6x + 8} > 1$$ holds if the exponent is negative: $$x^2 - 6x + 8 < 0.$$ Again factor: $$(x - 2)(x - 4) < 0,$$ which holds for $2 < x < 4$. Combining with $2 < x < 3$ from the base condition, we get $2 < x < 3$. 5. **Step 4: Combine all results** From the above, the solution set is $$x \\in (2, 3) \\cup (4, \\infty).$$ The smallest integral value satisfying this is $x = 5$. --- 6. **Problem 2: Find the number of solutions of the equation $$(2x - 3) 2^x = 1.$$** 7. **Step 1: Rewrite the equation** $$(2x - 3) 2^x = 1 \\Rightarrow 2x - 3 = 2^{-x}.$$ 8. **Step 2: Understand the functions** - $y_1 = 2x - 3$ is a straight line with slope 2 and y-intercept -3. - $y_2 = 2^{-x}$ is an exponential decay curve starting at $(0,1)$ and approaching 0 as $x \to \\infty$. 9. **Step 3: Find intersections** The number of solutions equals the number of intersection points of $y_1$ and $y_2$. 10. **Step 4: Analyze behavior** - At $x=0$, $y_1 = -3$, $y_2 = 1$, so $y_2 > y_1$. - At $x=2$, $y_1 = 1$, $y_2 = 2^{-2} = 0.25$, so $y_1 > y_2$. - At $x=1$, $y_1 = -1$, $y_2 = 0.5$, so $y_2 > y_1$. This suggests two points where the line crosses the curve. 11. **Final answer:** - The smallest integral $x$ satisfying the first inequality is $5$. - The equation $(2x - 3) 2^x = 1$ has **2 solutions**.