$9^3 \times (81)^2 \div (27)^3 = (3)^x$

### Concept & Formula This problem evaluates your ability to standardize bases and apply the core laws of indices to simplify algebraic expressions. $$ a^m \times a^n = a^{m+n} $$ $$ a^m \div a^n = a^{m-n} $$ $$ (a^m)^n = a^{mn} $$ ### Step-by-Step Solution * **Given:** The expression to evaluate is $9^3 \times (81)^2 \div (27)^3 = (3)^x$. * **Calculation:** 1. Observe that all numbers ($9$, $81$, $27$) are powers of the base $3$. Convert each term accordingly: $9 = 3^2$ $81 = 3^4$ $27 = 3^3$ 2. Substitute these prime bases back into the original equation: $(3^2)^3 \times (3^4)^2 \div (3^3)^3 = 3^x$ 3. Apply the power of a power rule to simplify the exponents: $3^6 \times 3^8 \div 3^9 = 3^x$ 4. Apply the product and quotient rules for exponents from left to right: $3^{(6 + 8 - 9)} = 3^x$ $3^{14 - 9} = 3^x$ $3^5 = 3^x$ 5. Since the bases on both sides are identical, equate the exponents: $x = 5$ ### Exam Strategy & Shortcut Do not write out full base conversions. Mentally translate the problem purely into the powers of base $3$. You know $9$ is $3^2$, so $9^3$ gives a power of $6$. $81$ is $3^4$, so squared it gives $8$. $27$ is $3^3$, so cubed it gives $9$. Combine them directly: $6 + 8 - 9 = 5$. The answer is $5$ in seconds. ### Common Pitfall Students often confuse the rules when raising a power to another power, mistakenly adding the exponents instead of multiplying them (e.g., computing $(3^2)^3$ as $3^5$ instead of $3^6$). Always multiply when brackets separate the powers. ### Final Answer **Therefore, the correct answer is 5.**