If a = 2 ln 2 , b = 3 ln 3 , and c = 5 ln 5 , compare the value of a , b , and c without using the calculator.
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Far simpler than using calculus!
Consider f ( x ) = x ln x . Then f ′ ( x ) = d x d f ( x ) = x 2 1 − ln x , ⟹ f ′ ( e ) = 0 . And f ′ ′ ( x ) = x 3 2 ln x − 3 , then f ′ ′ ( e ) < 0 . This implies that f ( x ) is maximum when x = e ≈ 3 . This means that f ( x ) decreases continuously from the maximum f ( e ) to f ( 3 ) , f ( 4 ) , f ( 5 ) ⋯ f ( ∞ ) . That is:
f ( 3 ) > 3 ln 3 > 3 ln 3 > ⟹ b > f ( 4 ) > f ( 5 ) 4 ln 4 > 5 ln 5 2 ln 2 > 5 ln 5 a > c
+1, very clever incorporation of the ln(2) term! I was going to use this method but since 2<e I abandoned it.
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Indeed. I look at the derivative and see the function's max is at (e) and think "That tells us f(3) > f(5), but how does that help us place f(2)?" This is a neat trick.
Same here!
3^10 > 2^15 > 5^6
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comparing a , b is the same as comparing 6 a , 6 b , that is comparing 3 ln ( 2 ) , 2 ln ( 3 ) = ln ( 8 ) , ln ( 9 ) . the latter is obviously bigger meaning b > a
comparing a , c is the same as comparing 1 0 a , 1 0 b , that is comparing 5 ln ( 2 ) , 2 ln ( 5 ) = ln ( 3 2 ) , ln ( 2 5 ) . the former is obviously bigger meaning a > c
compounding these we get b > a > c