Simplify: \( \left[\frac{3^{x+1}+12.3^{x-1}}{7.9^{x}}\right]^{\frac{1}{x}} \)
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To simplify the expression \( \left[\frac{3^{x+1}+12 \cdot 3^{x-1}}{7 \cdot 9^{x}}\right]^{\frac{1}{x}} \), we will start by simplifying the numerator and the denominator separately. First, let's simplify the numerator: \[ 3^{x+1} + 12 \cdot 3^{x-1} \] We can factor out \(3^{x-1}\) from both terms: \[ = 3^{x-1}(3^2 + 12) \] \[ = 3^{x-1}(9 + 12) = 3^{x-1}(21) = 21 \cdot 3^{x-1} \] Now, let’s simplify the denominator: \[ 7 \cdot 9^{x} = 7 \cdot (3^2)^x = 7 \cdot 3^{2x} \] Now substitute back into the expression: \[ \frac{3^{x+1} + 12 \cdot 3^{x-1}}{7 \cdot 9^{x}} = \frac{21 \cdot 3^{x-1}}{7 \cdot 3^{2x}} = \frac{21}{7} \cdot \frac{3^{x-1}}{3^{2x}} = 3 \cdot 3^{x-1 - 2x} = 3 \cdot 3^{-x - 1} = \frac{3}{3^{x+1}} = \frac{1}{3^x} \] Now we have: \[ \left[\frac{1}{3^x}\right]^{\frac{1}{x}} = (3^{-x})^{\frac{1}{x}} = 3^{-1} = \frac{1}{3} \] Thus, the simplified expression is: \[ \boxed{\frac{1}{3}} \]
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