Question:medium

\[ 9^{-z} = \frac{1}{27^x \cdot 27^y} = (81)^{-y} \]

Show Hint

Always convert all numbers to the same base before comparing exponents.
Updated On: Apr 18, 2026
  • (9/4, 9/8)
  • (3/2, 3/4)
  • (3,6)
  • (6,3)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
This problem involves indices (exponents). To find the values of \( x \) and \( y \), we should express all terms as powers of the same base, which is 3 in this case.
: Key Formula or Approach:
1. \( a^{mn} = (a^m)^n \).
2. \( \frac{1}{a^n} = a^{-n} \).
3. If \( a^x = a^y \), then \( x = y \) (for \( a>0, a \neq 1 \)).
Step 2: Detailed Explanation:
First, express the middle term as a power of 3: \[ \frac{1}{27\sqrt{27}} = \frac{1}{3^3 \cdot (3^3)^{1/2}} = \frac{1}{3^3 \cdot 3^{3/2}} = \frac{1}{3^{3 + 1.5}} = \frac{1}{3^{9/2}} = 3^{-9/2} \] Now, equate this to the first term: \[ 9^{-x} = (3^2)^{-x} = 3^{-2x} \] \[ 3^{-2x} = 3^{-9/2} \implies -2x = -\frac{9}{2} \implies x = \frac{9}{4} \] Next, equate it to the third term: \[ (81)^{-y} = (3^4)^{-y} = 3^{-4y} \] \[ 3^{-4y} = 3^{-9/2} \implies -4y = -\frac{9}{2} \implies y = \frac{9}{8} \] Therefore, the pair \( (x, y) = (9/4, 9/8) \).
Step 3: Final Answer:
The value of \( (x, y) \) is \( (9/4, 9/8) \).
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