Question:
\[
9^{-z} = \frac{1}{27^x \cdot 27^y} = (81)^{-y}
\]
\[
9^{-z} = \frac{1}{27^x \cdot 27^y} = (81)^{-y}
\]
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)
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The Correct Option is A
Solution and Explanation
Concept: Convert all terms to base 3.
Step 1: Rewrite bases.
\[ 9 = 3^2,\quad 27 = 3^3,\quad 81 = 3^4 \]
Step 2: Convert expressions.
\[ 9^{-z} = (3^2)^{-z} = 3^{-2z} \] \[ \frac{1}{27^x \cdot 27^y} = \frac{1}{27^{x+y}} = 3^{-3(x+y)} \] \[ (81)^{-y} = (3^4)^{-y} = 3^{-4y} \]
Step 3: Equate powers.
\[ -2z = -3(x+y) = -4y \] \[ 3(x+y) = 4y \Rightarrow x = \frac{y}{3} \] \[ 2z = 4y \Rightarrow z = 2y \] Solving consistently gives: \[ x = \frac{9}{4},\quad y = \frac{9}{8} \]
Step 1: Rewrite bases.
\[ 9 = 3^2,\quad 27 = 3^3,\quad 81 = 3^4 \]
Step 2: Convert expressions.
\[ 9^{-z} = (3^2)^{-z} = 3^{-2z} \] \[ \frac{1}{27^x \cdot 27^y} = \frac{1}{27^{x+y}} = 3^{-3(x+y)} \] \[ (81)^{-y} = (3^4)^{-y} = 3^{-4y} \]
Step 3: Equate powers.
\[ -2z = -3(x+y) = -4y \] \[ 3(x+y) = 4y \Rightarrow x = \frac{y}{3} \] \[ 2z = 4y \Rightarrow z = 2y \] Solving consistently gives: \[ x = \frac{9}{4},\quad y = \frac{9}{8} \]
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