To solve the equation \(2e^{3x} = 32\), we first isolate the exponential term:
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Divide both sides by 2: \[ e^{3x} = \frac{32}{2} = 16 \]
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Take the natural logarithm of both sides: \[ \ln(e^{3x}) = \ln(16) \]
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Using the property of logarithms that \(\ln(e^y) = y\): \[ 3x = \ln(16) \]
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Divide both sides by 3: \[ x = \frac{\ln(16)}{3} \]
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Now, calculate \(\ln(16)\): \[ \ln(16) = \ln(2^4) = 4\ln(2) \]
Using \(\ln(2) \approx 0.693147\): \[ \ln(16) \approx 4 \times 0.693147 \approx 2.772588 \]
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Now, substitute this value back to find \(x\): \[ x \approx \frac{2.772588}{3} \approx 0.924196 \]
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Rounding to the nearest two decimal places: \[ x \approx 0.92 \]
Thus, the answer is: \[ \boxed{0.92} \]