Egative power indicates an inverse

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The discussion explains the behavior of exponential functions for positive and negative values of x, specifically comparing 3^x, e^x, and 2^x. For x > 0, the inequality 3^x > e^x > 2^x holds true, while for x < 0, the reverse inequality 3^x < e^x < 2^x applies. This shift occurs because negative powers represent inverses, making smaller bases yield larger values when raised to negative exponents. The relationship is further clarified using logarithmic properties, indicating that if a < b, then 1/b < 1/a when both a and b are positive. Understanding these properties helps clarify the behavior of exponential functions across different ranges of x.
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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?
 
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e is approx. 2.72

so 3^1=3
2.72^1=2.72
2^1=2

therefore 3^x>e^x>2^x when x>0

however, when you get into the negatives

3^-1=1/3
2.72^-1=1/2.72
2^-1=1/2


this is because a negative power indicates an inverse( 1/x ), so the smaller the number with a negative power, the smaller the denomenator will be
 
gillgill said:
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.
 
All this is saying is that if 0<a< b< c then 0< 1/c< 1/b< 1/a

If a< b and a and b are positive, then 1< b/a because we have divided by a positive number.

Then 1/b< 1/a because we have divided by a positive number.
 
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