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

1. $-20$
2. $-12$
3. $-24$
4. $-16$

## 1 Answer

Given that,

$\left ( \sqrt{2} \right )^{19}3^{4}4^{2}9^{m}\cdot 8^{n} = 3^{n}\cdot 16^{m} \left ( 4\sqrt{64} \right )$

$\implies 2^{19/2}\cdot 3^{4}\cdot \left ( 2^{2} \right )^{2} \left ( 3^{2} \right )^{m} \left ( 2^{3} \right )^{n} = 3^{n} \left ( 2^{4} \right )^{m} \left ( 2^{6} \right )^{1/4}$

$\implies 2^{19/2}\cdot 3^{4}\cdot 2^{4}\cdot 3^{2m} \cdot 2^{3n} = 3^{n}\cdot 2^{4m}\cdot 2^{6/4}$

$\implies 2^{\left ( 19/2 + 4 + 3n \right )}\cdot 3^{\left ( 4 + 2m \right )} = 3^{n} \:2^{\left ( 4m + \frac{6}{4} \right )}$

Comparing both sides, we get

• $4 + 2m = n \qquad \qquad \to (i)$
• $\frac{19}{2} + 4 + 3n = 4m + \frac{3}{2} \qquad \to (ii)$

$(ii) \implies 3n+ \frac{19}{2}-\frac{3}{2}+4-4m=0$

$\implies 3n+ \frac{16}{2}+4-4m=0$

$\implies 3n+ 12-4m=0$

Put the value of $n$ from equation $(i),$ we get

$3\left ( 4+2m \right )+12-4m=0$

$\implies 12+6m+12-4m=0$

$\implies 2m + 24 =0$

$\implies 2m = -24$

$\implies m = -\frac{24}{2}$

$\boxed{m = -12}$

Correct Answer$: \text{B}$

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