Using properties of exponents: $$\frac{2^{-3} \cdot 2^7}{2^{-1} \cdot 2^2} = \frac{2^{-3+7}}{2^{-1+2}} = \frac{2^4}{2^1} = 2^{4-1} = 2^3$$. Choice B results from incorrectly computing the numerator as $$2^{-3-7}$$. Choice C comes from adding all exponents instead of applying quotient rules. Choice D results from computing $$2^{-11}$$ by incorrectly subtracting exponents.

Using the product rule: $$5^{-2} \cdot 5^n \cdot 5^{-3} = 5^{-2+n+(-3)} = 5^{n-5}$$. For this to equal $$5^2$$, we need $$n-5 = 2$$, so $$n = 7$$. Choice A gives $$5^{-3}$$, choice C gives $$5^{-2}$$, and choice D gives $$5^{-8}$$.

First, $$(2^{-3})^{-2} = 2^{(-3)(-2)} = 2^6$$. The reciprocal of $$2^6$$ is $$\frac{1}{2^6} = 2^{-6}$$. Choice B is the original expression, not its reciprocal. Choice C equals $$2^{-6}$$ but isn't simplified. Choice D equals $$2^6$$, which is the original expression.

When you encounter expressions with exponents that need simplifying, your goal is to use the rules of exponents to combine like terms. This problem tests your understanding of the power rule, product rule, and quotient rule for exponents.

Let's work through this step-by-step. First, apply the power rule to $$(3^{-2})^4$$. When raising a power to a power, you multiply the exponents: $$(3^{-2})^4 = 3^{-2 \cdot 4} = 3^{-8}$$.

Now your expression becomes: $$\frac{3^{-8} \cdot 3^{-1}}{3^{-5} \cdot 3^{-4}}$$

Next, use the product rule in both the numerator and denominator. When multiplying powers with the same base, you add the exponents:

• Numerator: $$3^{-8} \cdot 3^{-1} = 3^{-8 + (-1)} = 3^{-9}$$
• Denominator: $$3^{-5} \cdot 3^{-4} = 3^{-5 + (-4)} = 3^{-9}$$

This gives you: $$\frac{3^{-9}}{3^{-9}} = 3^{-9 - (-9)} = 3^{-9 + 9} = 3^0 = 1$$

Answer A ($$3^{-8}$$) comes from stopping after the first step and forgetting to continue simplifying. Answer B ($$3^{-18}$$) results from incorrectly multiplying all the exponents together instead of using the proper rules. Answer C ($$\frac{1}{3}$$) might come from calculation errors when combining exponents.

Remember: when working with exponent problems, apply the rules systematically—power rule first, then product/quotient rules. Double-check your arithmetic when adding and subtracting negative exponents, as sign errors are common here.

$$\frac{6^{-2} \cdot 6^{-4}}{6^{-8}} = \frac{6^{-6}}{6^{-8}} = 6^{-6-(-8)} = 6^{2}$$. Choice A results from multiplying all exponents. Choice C results from adding all exponents. Choice D represents $$6^{-2}$$, which would result from computing $$6^{-6+(-8)}$$.

Substituting: $$\frac{x^2}{y} = \frac{(3^{-2})^2}{3^{-4}} = \frac{3^{-4}}{3^{-4}} = 3^{-4-(-4)} = 3^0 = 1$$. Choice B results from multiplying exponents instead of using the power rule correctly. Choice C comes from computing $$3^{-4+4}$$ but forgetting the power of 2. Choice D results from computing $$3^{(-2)^2 - (-4)} = 3^{4+4}$$, incorrectly applying the power rule.

$$\frac{1}{64} = \frac{1}{4^3} = 4^{-3}$$ (choice A), $$\frac{1}{64} = \frac{1}{2^6} = 2^{-6}$$ (choice B), and $$(\frac{1}{4})^3 = \frac{1}{4^3} = \frac{1}{64}$$ (choice C). However, $$8^{-3} = \frac{1}{8^3} = \frac{1}{512}$$, which is not equal to $$\frac{1}{64}$$. Students might confuse $$8^3 = 512$$ with $$4^3 = 64$$.

When you see exponents being divided, you need to apply the quotient rule for exponents, which states that $$\frac{x^a}{x^b} = x^{a-b}$$. The key is to subtract the bottom exponent from the top exponent.

Let's work through this problem correctly. You have $$\frac{x^{-4}}{x^{-7}}$$, so you subtract the exponents: $$x^{-4-(-7)} = x^{-4+7} = x^3$$. The correct answer should be $$x^3$$, not $$x^{-28}$$.

The student got $$x^{-28}$$ by multiplying the exponents: $$(-4) \times(-7) = 28$$, then keeping it negative as $$x^{-28}$$. This is exactly what answer choice C describes - multiplying exponents instead of using the quotient rule correctly.

Looking at the other options: A is wrong because the issue isn't about the base $$x$$ or whether the final exponent should be positive or negative. B is incorrect because adding the exponents ($$-4 + (-7) = -11$$) would give you $$x^{-11}$$, which is what you'd do when multiplying terms with the same base, not dividing them. D doesn't make sense because there's no "power rule for negative exponents" being applied here - this is purely about the quotient rule.

Remember this pattern: when dividing powers with the same base, always subtract the exponents (bottom from top). When multiplying powers with the same base, you add the exponents. Don't let negative signs confuse you - just follow the subtraction rule carefully.

When you encounter expressions with exponents that need to be simplified, your goal is to use the rules of exponents to combine terms and then substitute the given value.

Let's work through this step-by-step using the exponent rules. First, simplify $$\frac{4^{-2} \cdot 4^{x}}{4^{-3}}$$ before substituting $$x = -1$$.

Using the product rule, $$4^{-2} \cdot 4^{x} = 4^{-2+x}$$. Then using the quotient rule, $$\frac{4^{-2+x}}{4^{-3}} = 4^{(-2+x)-(-3)} = 4^{-2+x+3} = 4^{1+x}$$.

Now substitute $$x = -1$$: $$4^{1+(-1)} = 4^{0} = 1$$. Since any non-zero number raised to the zero power equals 1, the expression equals $$4^{0}$$ with a value of 1.

Looking at the wrong answers: Answer A incorrectly calculates the exponent as 2, perhaps by misapplying the quotient rule or making sign errors. Answer B gets $$4^{-6}$$, which might result from adding all the exponents incorrectly ($$-2 + (-1) + (-3) = -6$$) instead of properly applying the quotient rule. Answer D gives $$4^{-2}$$, possibly from incorrectly combining the original exponents or stopping the calculation too early.

The key strategy here is to always simplify expressions with exponents algebraically first, then substitute the variable's value. This approach prevents calculation errors and makes the work cleaner. Remember: when dividing powers with the same base, subtract the exponents, and when multiplying, add them.

Since $$a^{-3} = \frac{1}{8}$$, we have $$a^3 = 8$$, so $$a = 2$$. Therefore, $$a^{-6} = 2^{-6} = \frac{1}{2^6} = \frac{1}{64}$$. Alternatively, $$a^{-6} = (a^{-3})^2 = (\frac{1}{8})^2 = \frac{1}{64}$$. Choice A represents $$a^{-4}$$, choice C represents $$a^{-9}$$, and choice D represents $$a^{-12}$$.