If $m$ and $n$ are integers such that $(\sqrt{2})^{19} 3^{4} 4^{2} 9^m8^n = 3^n 16^m (\sqrt[4]{64})$ then $m$ is

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

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Convert everything to prime factors (2 and 3) because comparing powers of the same base is the easiest way to solve these equations.

Left side breakdown:

• $(\sqrt{2})^{19} = (2^{1/2})^{19} = 2^{19/2}$
• $3^4 = 3^4$
• $4^2 = (2^2)^2 = 2^4$
• $9^m = (3^2)^m = 3^{2m}$
• $8^n = (2^3)^n = 2^{3n}$

Right side breakdown:

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

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Combine powers of same bases:

Left side:

• Powers of 2: $2^{19/2} \cdot 2^4 \cdot 2^{3n} = 2^{19/2 + 4 + 3n} = 2^{27/2 + 3n}$
• Powers of 3: $3^4 \cdot 3^{2m} = 3^{4 + 2m}$

Right side:

• Powers of 2: $2^{4m} \cdot 2^{3/2} = 2^{4m + 3/2}$
• Powers of 3: $3^n$

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Since both sides must be equal, the powers of 2 must match and powers of 3 must match.

For powers of 2:

$\dfrac{27}{2} + 3n = 4m + \dfrac{3}{2}$ ... (1)

For powers of 3: $4 + 2m = n$ ... (2)

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From equation (2): $n = 4 + 2m$

Substitute into equation (1):

$\dfrac{27}{2} + 3(4 + 2m) = 4m + \dfrac{3}{2}$

$\dfrac{27}{2} + 12 + 6m = 4m + \dfrac{3}{2}$

$\dfrac{27}{2} - \dfrac{3}{2} + 12 = 4m - 6m$

$\dfrac{24}{2} + 12 = -2m$

$12 + 12 = -2m$

$24 = -2m$

$m = -12$

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Verify: If $m = -12$, then $n = 4 + 2(-12) = -20$

Check equation (1):

$\dfrac{27}{2} + 3(-20) = 4(-12) + \dfrac{3}{2}$

$\dfrac{27}{2} - 60 = -48 + \dfrac{3}{2}$

$\dfrac{-93}{2} = \dfrac{-93}{2}$ ✓

Therefore, $m = -12$