Let the two powers be \( x \) and \( y \). According to the problem, we have two equations based on the product and quotient:
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The product of \( x \) and \( y \) is given by: \[ x \cdot y = 4^{12} \]
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The quotient of \( x \) and \( y \) is given by: \[ \frac{x}{y} = 4^6 \]
To express \( x \) and \( y \) in terms of powers of 4, we can rewrite them as follows:
- Let \( x = 4^a \)
- Let \( y = 4^b \)
Now substituting these into our equations gives us:
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From the product: \[ (4^a)(4^b) = 4^{12} \] Using the properties of exponents, this can be expressed as: \[ 4^{a+b} = 4^{12} \] Thus, we have: \[ a + b = 12 \quad \text{(1)} \]
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From the quotient: \[ \frac{4^a}{4^b} = 4^6 \] Again using the properties of exponents: \[ 4^{a-b} = 4^6 \] Therefore, we get: \[ a - b = 6 \quad \text{(2)} \]
Now we need to solve the system of equations formed by (1) and (2):
From equation (2): \[ a - b = 6 \implies a = b + 6 \]
Substituting \( a \) in equation (1): \[ (b + 6) + b = 12 \] This simplifies to: \[ 2b + 6 = 12 \] Subtracting 6 from both sides gives: \[ 2b = 6 \] Dividing by 2: \[ b = 3 \]
Now substituting \( b \) back to find \( a \): \[ a = b + 6 = 3 + 6 = 9 \]
Thus, we have: \[ a = 9 \quad \text{and} \quad b = 3 \]
This means: \[ x = 4^9 \quad \text{and} \quad y = 4^3 \]
Verification
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Verifying the product: \[ x \cdot y = 4^9 \cdot 4^3 = 4^{9+3} = 4^{12} \] This is correct.
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Verifying the quotient: \[ \frac{x}{y} = \frac{4^9}{4^3} = 4^{9-3} = 4^6 \] This is also correct.
In conclusion, the two powers are: \[ x = 4^9 \quad \text{and} \quad y = 4^3 \]