gillgill
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Why is 3^x > e^x > 2^x when x>0, but 3^x < e^x < 2^x when x<0?
The discussion revolves around the behavior of exponential functions with different bases, specifically comparing \(3^x\), \(e^x\), and \(2^x\) for positive and negative values of \(x\). Participants explore why the inequalities change based on the sign of \(x\).
The discussion is active, with participants providing insights into the behavior of the functions and exploring different perspectives on the inequalities. There is no explicit consensus, but various interpretations and reasoning approaches are being examined.
Participants reference the approximate value of \(e\) and discuss the implications of negative powers without resolving the broader implications of these observations.
Another way of looking at it. Since 3 > e and therefore Log(3) > 1, x Log(3) > x implies x > 0 and x Log(3) < x implies x < 0.gillgill said:Why is 3^x > e^x > 2^x when x>0, but 3^x < e^x < 2^x when x<0?