Given $$a = 2^3 = 8$$ and $$b = 2^4 = 16$$. We need to find $$\frac{a^2 \cdot b^3}{(ab)^2}$$. Using exponent rules: $$\frac{a^2 \cdot b^3}{(ab)^2} = \frac{a^2 \cdot b^3}{a^2 \cdot b^2} = \frac{b^3}{b^2} = b^1 = b = 2^4 = 16$$. Alternatively, substituting the values directly: $$a^2 = (2^3)^2 = 2^6 = 64$$, $$b^3 = (2^4)^3 = 2^{12} = 4096$$, and $$(ab)^2 = (2^3 \cdot 2^4)^2 = (2^7)^2 = 2^{14}$$. So $$\frac{a^2 \cdot b^3}{(ab)^2} = \frac{2^6 \cdot 2^{12}}{2^{14}} = \frac{2^{18}}{2^{14}} = 2^4 = 16$$. Choice A (4) would result from incorrectly calculating $$b^3/b^2 = b$$ but using $$a$$ instead of $$b$$. Choice B (8) might come from errors in exponent arithmetic. Choice D (32) could result from miscalculating $$2^4$$ as $$2^5$$.

First, simplify the numerator: $$(3^2)^x \cdot 3^{-x} = 3^{2x} \cdot 3^{-x} = 3^{2x-x} = 3^x$$. So the expression becomes $$\frac{3^x}{3^{x-1}}$$. Using the quotient rule: $$3^{x-(x-1)} = 3^{x-x+1} = 3^1 = 3$$. The expression simplifies to the constant 3, regardless of the value of $$x$$. Choice B (9) might result from incorrectly calculating $$3^1$$ as $$3^2$$. Choice C ($$\frac{1}{3}$$) could come from sign errors in the exponent arithmetic. Choice D ($$3^x$$) would result from incorrectly simplifying the quotient rule, perhaps by subtracting exponents in the wrong direction or forgetting to apply the rule entirely.

First, simplify the expression inside the parentheses: $$\frac{2^3}{2^{-1}} = 2^3 \cdot 2^1 = 2^{3-(-1)} = 2^{3+1} = 2^4$$. Then: $$\left(\frac{2^3}{2^{-1}}\right)^2 = (2^4)^2 = 2^8$$. Finally: $$2^8 \div 2^4 = 2^{8-4} = 2^4$$. Choice B ($$2^6$$) might result from incorrectly calculating $$2^8 \div 2^4$$ as $$2^{8-2}$$. Choice C ($$2^8$$) would result from forgetting to divide by $$2^4$$ at the end. Choice D ($$2^{12}$$) could come from incorrectly multiplying exponents instead of using the quotient rule in the final step.

First, simplify the left side: $$(2^k)^3 \cdot 2^{-5} = 2^{3k} \cdot 2^{-5} = 2^{3k-5}$$. Since $$\frac{1}{4} = \frac{1}{2^2} = 2^{-2}$$, we have the equation $$2^{3k-5} = 2^{-2}$$. Therefore $$3k - 5 = -2$$, which gives us $$3k = 3$$, so $$k = 1$$. Let's verify: when $$k = 1$$, $$(2^1)^3 \cdot 2^{-5} = 2^3 \cdot 2^{-5} = 2^{-2} = \frac{1}{4}$$ ✓. Choice A would give $$(2^{-1})^3 \cdot 2^{-5} = 2^{-3} \cdot 2^{-5} = 2^{-8} = \frac{1}{256}$$. Choice B would give $$(2^0)^3 \cdot 2^{-5} = 1 \cdot 2^{-5} = \frac{1}{32}$$. Choice D would give $$(2^2)^3 \cdot 2^{-5} = 2^6 \cdot 2^{-5} = 2^1 = 2$$.