Compare an expression with zero
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I need to compare $1-frac{2}{3}cdot3^{-frac{2}{3}}cdot log_e9$ and $0$ without any computer
logarithms
asked Nov 15 at 11:57
Марк Тюков
103
add a comment |
up vote
0
down vote
favorite
I need to compare $1-frac{2}{3}cdot3^{-frac{2}{3}}cdot log_e9$ and $0$ without any computer
logarithms
asked Nov 15 at 11:57
Марк Тюков
103
You just need to see if it's positive or not. After some fiddling, you'll see that this is equivalent to seeing if $displaystylelog_e9 < frac{3^frac{5}{3}}{2}$.
– user3482749
Nov 15 at 12:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to compare $1-frac{2}{3}cdot3^{-frac{2}{3}}cdot log_e9$ and $0$ without any computer
logarithms
asked Nov 15 at 11:57
Марк Тюков
103
I need to compare $1-frac{2}{3}cdot3^{-frac{2}{3}}cdot log_e9$ and $0$ without any computer
logarithms
logarithms
asked Nov 15 at 11:57
Марк Тюков
103
asked Nov 15 at 11:57
Марк Тюков
103
asked Nov 15 at 11:57
Марк Тюков
103
asked Nov 15 at 11:57
Марк Тюков
103
asked Nov 15 at 11:57
Марк Тюков
103
103
You just need to see if it's positive or not. After some fiddling, you'll see that this is equivalent to seeing if $displaystylelog_e9 < frac{3^frac{5}{3}}{2}$.
– user3482749
Nov 15 at 12:02
add a comment |
You just need to see if it's positive or not. After some fiddling, you'll see that this is equivalent to seeing if $displaystylelog_e9 < frac{3^frac{5}{3}}{2}$.
– user3482749
Nov 15 at 12:02
You just need to see if it's positive or not. After some fiddling, you'll see that this is equivalent to seeing if $displaystylelog_e9 < frac{3^frac{5}{3}}{2}$.
– user3482749
Nov 15 at 12:02
You just need to see if it's positive or not. After some fiddling, you'll see that this is equivalent to seeing if $displaystylelog_e9 < frac{3^frac{5}{3}}{2}$.
– user3482749
Nov 15 at 12:02
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
With the use of $;9<e^3$ we get
$$frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9<frac{2}{3}cdot3^{-frac{2}{3}}cdot ln e^3=frac{2}{9^{1over 3}}<frac{2}{8^{1over 3}}=1,$$ thus $$1-frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9>0.$$
answered Nov 15 at 12:15
user376343
2,4581718
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up vote
0
down vote
It's easy to take conservative estimates $3^{-frac{2}{3}}<frac{1}{2}$, $log_e9<frac{5}{2}$ and conclude that the expression is positive.
answered Nov 15 at 12:12
Vasya
3,2791515
And why is $log_e 3$ less than $frac{5}{2}$?
– Марк Тюков
Nov 15 at 12:22
@МаркТюков: because $eapprox 2.7$, $e^2approx 7.3$, $sqrt{e} > 1.5$
– Vasya
Nov 15 at 12:41
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
With the use of $;9<e^3$ we get
$$frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9<frac{2}{3}cdot3^{-frac{2}{3}}cdot ln e^3=frac{2}{9^{1over 3}}<frac{2}{8^{1over 3}}=1,$$ thus $$1-frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9>0.$$
answered Nov 15 at 12:15
user376343
2,4581718
add a comment |
up vote
0
down vote
accepted
With the use of $;9<e^3$ we get
$$frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9<frac{2}{3}cdot3^{-frac{2}{3}}cdot ln e^3=frac{2}{9^{1over 3}}<frac{2}{8^{1over 3}}=1,$$ thus $$1-frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9>0.$$
answered Nov 15 at 12:15
user376343
2,4581718
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
With the use of $;9<e^3$ we get
$$frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9<frac{2}{3}cdot3^{-frac{2}{3}}cdot ln e^3=frac{2}{9^{1over 3}}<frac{2}{8^{1over 3}}=1,$$ thus $$1-frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9>0.$$
answered Nov 15 at 12:15
user376343
2,4581718
With the use of $;9<e^3$ we get
$$frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9<frac{2}{3}cdot3^{-frac{2}{3}}cdot ln e^3=frac{2}{9^{1over 3}}<frac{2}{8^{1over 3}}=1,$$ thus $$1-frac{2}{3}cdot3^{-frac{2}{3}}cdot ln 9>0.$$
answered Nov 15 at 12:15
user376343
2,4581718
edited Nov 15 at 12:20
answered Nov 15 at 12:15
user376343
2,4581718
answered Nov 15 at 12:15
user376343
2,4581718
answered Nov 15 at 12:15
user376343
2,4581718
2,4581718
add a comment |
add a comment |
up vote
0
down vote
It's easy to take conservative estimates $3^{-frac{2}{3}}<frac{1}{2}$, $log_e9<frac{5}{2}$ and conclude that the expression is positive.
answered Nov 15 at 12:12
Vasya
3,2791515
And why is $log_e 3$ less than $frac{5}{2}$?
– Марк Тюков
Nov 15 at 12:22
@МаркТюков: because $eapprox 2.7$, $e^2approx 7.3$, $sqrt{e} > 1.5$
– Vasya
Nov 15 at 12:41
add a comment |
up vote
0
down vote
It's easy to take conservative estimates $3^{-frac{2}{3}}<frac{1}{2}$, $log_e9<frac{5}{2}$ and conclude that the expression is positive.
answered Nov 15 at 12:12
Vasya
3,2791515
And why is $log_e 3$ less than $frac{5}{2}$?
– Марк Тюков
Nov 15 at 12:22
@МаркТюков: because $eapprox 2.7$, $e^2approx 7.3$, $sqrt{e} > 1.5$
– Vasya
Nov 15 at 12:41
add a comment |
up vote
0
down vote
up vote
0
down vote
It's easy to take conservative estimates $3^{-frac{2}{3}}<frac{1}{2}$, $log_e9<frac{5}{2}$ and conclude that the expression is positive.
answered Nov 15 at 12:12
Vasya
3,2791515
It's easy to take conservative estimates $3^{-frac{2}{3}}<frac{1}{2}$, $log_e9<frac{5}{2}$ and conclude that the expression is positive.
answered Nov 15 at 12:12
Vasya
3,2791515
answered Nov 15 at 12:12
Vasya
3,2791515
answered Nov 15 at 12:12
Vasya
3,2791515
answered Nov 15 at 12:12
Vasya
3,2791515
3,2791515
And why is $log_e 3$ less than $frac{5}{2}$?
– Марк Тюков
Nov 15 at 12:22
@МаркТюков: because $eapprox 2.7$, $e^2approx 7.3$, $sqrt{e} > 1.5$
– Vasya
Nov 15 at 12:41
add a comment |
And why is $log_e 3$ less than $frac{5}{2}$?
– Марк Тюков
Nov 15 at 12:22
@МаркТюков: because $eapprox 2.7$, $e^2approx 7.3$, $sqrt{e} > 1.5$
– Vasya
Nov 15 at 12:41
And why is $log_e 3$ less than $frac{5}{2}$?
– Марк Тюков
Nov 15 at 12:22
And why is $log_e 3$ less than $frac{5}{2}$?
– Марк Тюков
Nov 15 at 12:22
@МаркТюков: because $eapprox 2.7$, $e^2approx 7.3$, $sqrt{e} > 1.5$
– Vasya
Nov 15 at 12:41
@МаркТюков: because $eapprox 2.7$, $e^2approx 7.3$, $sqrt{e} > 1.5$
– Vasya
Nov 15 at 12:41
add a comment |
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You just need to see if it's positive or not. After some fiddling, you'll see that this is equivalent to seeing if $displaystylelog_e9 < frac{3^frac{5}{3}}{2}$.
– user3482749
Nov 15 at 12:02