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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???
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???