Let $r$ be the result of doubling both the base and the exponent of $a^b$, $b \neq 0$. If $r$ equals the product of $a^b$ by $x^b$, then $x$ equals

### Concept & Formula This problem tests the fundamental rules of exponents, specifically the power of a product rule and how to equate expressions with the same exponent. The core principle is recognizing how changes to the base and exponent affect the algebraic expression. $$(xy)^n = x^n \cdot y^n$$ ### Step-by-Step Solution * **Given:** The original term is $a^b$. $r$ is the result of doubling the base (from $a$ to $2a$) and the exponent (from $b$ to $2b$). Also, $r = a^b \cdot x^b$. * **Calculation / Deduction:** 1. Express $r$ algebraically based on the first condition: $$r = (2a)^{2b}$$ 2. Express the second condition mathematically: $$r = a^b \cdot x^b$$ 3. Combine the two equations: $$(2a)^{2b} = a^b \cdot x^b$$ 4. Simplify the RHS using the rule $(xy)^n = x^n \cdot y^n$: $$(2a)^{2b} = (ax)^b$$ 5. To compare bases, we need the exponents to match. Rewrite the LHS so it has an exponent of $b$: $$((2a)^2)^b = (ax)^b$$ 6. Expand the inner term on the LHS: $$(4a^2)^b = (ax)^b$$ 7. Since the exponents are identical and non-zero ($b \neq 0$), we can equate the bases: $$4a^2 = ax$$ 8. Solve for $x$: $$x = \frac{4a^2}{a} = 4a$$ ### Exam Strategy & Shortcut **Value Assumption (Plug-in Method):** Let $a = 2$ and $b = 1$. Original term $= 2^1 = 2$. Double base ($2 \rightarrow 4$) and exponent ($1 \rightarrow 2$). So, $r = 4^2 = 16$. Given $r = a^b \cdot x^b \implies 16 = 2^1 \cdot x^1 \implies 2x = 16 \implies x = 8$. Now check the options with $a = 2$: (a) 2 (b) 4 (c) $2a = 2(2) = 4$ (d) $4a = 4(2) = 8$ Option (d) yields 8, perfectly matching our result. ### Common Pitfall The most frequent error is misinterpreting "doubling both the base and the exponent". Students often write $2(a^b)$ or $(2a)^b$ instead of properly squaring the entire new base and doubling the power to yield $(2a)^{2b}$. ### Final Answer **Therefore, the correct answer is $4a$.**