A question about exponential functions
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How can we prove that for b>a>e (e being the euler’s number), a^b is greater than b^a?
exponential-function eulers-constant
asked Nov 28 at 22:08
S.Esk
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How can we prove that for b>a>e (e being the euler’s number), a^b is greater than b^a?
exponential-function eulers-constant
asked Nov 28 at 22:08
S.Esk
33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can we prove that for b>a>e (e being the euler’s number), a^b is greater than b^a?
exponential-function eulers-constant
asked Nov 28 at 22:08
S.Esk
33
How can we prove that for b>a>e (e being the euler’s number), a^b is greater than b^a?
exponential-function eulers-constant
exponential-function eulers-constant
asked Nov 28 at 22:08
S.Esk
33
asked Nov 28 at 22:08
S.Esk
33
asked Nov 28 at 22:08
S.Esk
33
asked Nov 28 at 22:08
S.Esk
33
asked Nov 28 at 22:08
S.Esk
33
33
add a comment |
add a comment |
1 Answer
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You want to prove $x^{1/x}$, or equivalently its logarithm $x^{-1}ln x$, decreases in $xge e$ so that $a^{1/a}>b^{1/b}$. Use the product rule.
answered Nov 28 at 22:11
J.G.
21.3k21933
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
up vote
0
down vote
accepted
You want to prove $x^{1/x}$, or equivalently its logarithm $x^{-1}ln x$, decreases in $xge e$ so that $a^{1/a}>b^{1/b}$. Use the product rule.
answered Nov 28 at 22:11
J.G.
21.3k21933
add a comment |
up vote
0
down vote
accepted
You want to prove $x^{1/x}$, or equivalently its logarithm $x^{-1}ln x$, decreases in $xge e$ so that $a^{1/a}>b^{1/b}$. Use the product rule.
answered Nov 28 at 22:11
J.G.
21.3k21933
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You want to prove $x^{1/x}$, or equivalently its logarithm $x^{-1}ln x$, decreases in $xge e$ so that $a^{1/a}>b^{1/b}$. Use the product rule.
answered Nov 28 at 22:11
J.G.
21.3k21933
You want to prove $x^{1/x}$, or equivalently its logarithm $x^{-1}ln x$, decreases in $xge e$ so that $a^{1/a}>b^{1/b}$. Use the product rule.
answered Nov 28 at 22:11
J.G.
21.3k21933
answered Nov 28 at 22:11
J.G.
21.3k21933
answered Nov 28 at 22:11
J.G.
21.3k21933
answered Nov 28 at 22:11
J.G.
21.3k21933
21.3k21933
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