Question

If m and n are integers such that \((\sqrt{2})^{19} 3^{4} 4^{2} 9^{m} 8^{n}= 3^{n} 16^{m} (^4\sqrt{64})\) then m is

• A. -20
• B. -12
• C. -24
• D. -16

Answer 1

The Correct Option is B

Given: \[ (\sqrt{2})^{19} \cdot 3^4 \cdot 4^2 \cdot 9^m \cdot 8^n = 3^n \cdot 16^m \cdot \sqrt[4]{64} \]

Step 1: Express all components in base primes

• \( \sqrt{2} = 2^{1/2} \Rightarrow (\sqrt{2})^{19} = 2^{19/2} \)
• \( 4^2 = (2^2)^2 = 2^4 \)
• \( 9^m = (3^2)^m = 3^{2m} \)
• \( 8^n = (2^3)^n = 2^{3n} \)
• \( 16^m = (2^4)^m = 2^{4m} \)
• \( \sqrt[4]{64} = (2^6)^{1/4} = 2^{3/2} \)

Substitute back into the equation: \[ 2^{19/2} \cdot 2^4 \cdot 2^{3n} \cdot 3^4 \cdot 3^{2m} = 2^{4m} \cdot 2^{3/2} \cdot 3^n \]

Step 2: Combine exponents

Left side: \[ 2^{19/2 + 4 + 3n} \cdot 3^{4 + 2m} \] Right side: \[ 2^{4m + 3/2} \cdot 3^n \]

Step 3: Equate powers of same base

• Base 2: \( \frac{19}{2} + 4 + 3n = 4m + \frac{3}{2} \)
• Base 3: \( 4 + 2m = n \)

Step 4: Solve the system of equations

From base 3: \[ n = 4 + 2m \quad \text{(i)} \] Substitute into base 2 equation: \[ \frac{19}{2} + 4 + 3(4 + 2m) = 4m + \frac{3}{2} \] \[ \frac{27}{2} + 12 + 6m = 4m + \frac{3}{2} \] \[ \frac{51}{2} + 6m = 4m + \frac{3}{2} \] Multiply both sides by 2: \[ 51 + 12m = 8m + 3 \Rightarrow 4m = -48 \Rightarrow m = -12 \]

Final Answer:

\[ \boxed{m = -12} \]