Show that $19^{31}>13^{33}$If $displaystyle x= frac{e^{3z}}{y^4}$ then $z(x,y)$?Algebra and exponentsI'm...
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Show that $19^{31}>13^{33}$
If $displaystyle x= frac{e^{3z}}{y^4}$ then $z(x,y)$?Algebra and exponentsI'm stuck with this summation problemshow a function remains positive for a given domain$b>a>0$ , how to prove that $Big(dfrac {sqrt a +sqrt b}{2}Big )^2 < dfrac 1e Big(dfrac {b^b} {a^a}Big)^{dfrac 1{b-a}}$?Simplify $(frac{x+5}{x^2-81} + frac{x+7}{x^2-18x+81}):(frac{x+3}{x-9})^2 + frac{7+x}{9+x}$Which one is bigger? Logarithms and trigonometric functionsFinding $ lim_{xrightarrow 0}frac{(1+x)^{frac{1}{x}}-e+frac{ex}{2}}{x^2}$ without using series expansion.maximum value of expression $(sqrt{-3+4x-x^2}+4)^2+(x-5)^2$Coefficient of $x^{n-3}$ in $prod^{n}_{k=1}(x-k)$
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How can i prove that $19^{31}>13^{33}$?
What I tried
$$bigg(frac{19}{13}bigg)^2=frac{361}{169}>2>1$$
then $19^{2}>13^{2}$ and $displaystyle 19^{30}>13^{30}$
How do I show it. Help me please.
algebra-precalculus exponential-function number-comparison
edited 11 hours ago
Michael Rozenberg
107k1894198
asked 15 hours ago
jackyjacky
960615
$endgroup$
add a comment |
$begingroup$
How can i prove that $19^{31}>13^{33}$?
What I tried
$$bigg(frac{19}{13}bigg)^2=frac{361}{169}>2>1$$
then $19^{2}>13^{2}$ and $displaystyle 19^{30}>13^{30}$
How do I show it. Help me please.
algebra-precalculus exponential-function number-comparison
edited 11 hours ago
Michael Rozenberg
107k1894198
asked 15 hours ago
jackyjacky
960615
$endgroup$
add a comment |
$begingroup$
How can i prove that $19^{31}>13^{33}$?
What I tried
$$bigg(frac{19}{13}bigg)^2=frac{361}{169}>2>1$$
then $19^{2}>13^{2}$ and $displaystyle 19^{30}>13^{30}$
How do I show it. Help me please.
algebra-precalculus exponential-function number-comparison
edited 11 hours ago
Michael Rozenberg
107k1894198
asked 15 hours ago
jackyjacky
960615
$endgroup$
How can i prove that $19^{31}>13^{33}$?
What I tried
$$bigg(frac{19}{13}bigg)^2=frac{361}{169}>2>1$$
then $19^{2}>13^{2}$ and $displaystyle 19^{30}>13^{30}$
How do I show it. Help me please.
algebra-precalculus exponential-function number-comparison
algebra-precalculus exponential-function number-comparison
edited 11 hours ago
Michael Rozenberg
107k1894198
asked 15 hours ago
jackyjacky
960615
edited 11 hours ago
Michael Rozenberg
107k1894198
asked 15 hours ago
jackyjacky
960615
edited 11 hours ago
Michael Rozenberg
107k1894198
edited 11 hours ago
Michael Rozenberg
107k1894198
edited 11 hours ago
Michael Rozenberg
107k1894198
107k1894198
asked 15 hours ago
jackyjacky
960615
asked 15 hours ago
jackyjacky
960615
asked 15 hours ago
jackyjacky
960615
960615
add a comment |
add a comment |
7 Answers
7
active
oldest
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$begingroup$
Note that by the binomial theorem
$$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
which implies that $19^{31}>13^{33}$.
Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?
answered 15 hours ago
Robert ZRobert Z
99.8k1068140
$endgroup$
$begingroup$
Solved bonus below(before seeing it posted).
$endgroup$
– miniparser
14 hours ago
add a comment |
$begingroup$
$$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$
$$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$
Clearly $$2^{15}geq 13cdot 9implies A>B$$
answered 15 hours ago
greedoidgreedoid
46.1k1160117
$endgroup$
add a comment |
$begingroup$
Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
$x ln(19) > (x+2) ln(13)$
and finally
$ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.
answered 15 hours ago
Gâteau-GalloisGâteau-Gallois
410213
$endgroup$
add a comment |
$begingroup$
By your work
$$frac{19^2}{13^2}>2,$$ which says
$$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!
By this way we can prove that
$$19^{31}>13^{35}.$$
answered 11 hours ago
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
$19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$
$6859gt 3times 2197$
$6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$
$27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$
answered 14 hours ago
miniparserminiparser
7871616
$endgroup$
add a comment |
$begingroup$
Well, using what you've got so far:
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.
$19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.
=====
For what it's worth one way I might do it if it were up to me is to note.
$19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.
So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.
answered 10 hours ago
fleabloodfleablood
71.9k22687
$endgroup$
add a comment |
$begingroup$
$left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$
answered 7 hours ago
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that by the binomial theorem
$$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
which implies that $19^{31}>13^{33}$.
Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?
answered 15 hours ago
Robert ZRobert Z
99.8k1068140
$endgroup$
$begingroup$
Solved bonus below(before seeing it posted).
$endgroup$
– miniparser
14 hours ago
add a comment |
$begingroup$
Note that by the binomial theorem
$$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
which implies that $19^{31}>13^{33}$.
Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?
answered 15 hours ago
Robert ZRobert Z
99.8k1068140
$endgroup$
$begingroup$
Solved bonus below(before seeing it posted).
$endgroup$
– miniparser
14 hours ago
add a comment |
$begingroup$
Note that by the binomial theorem
$$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
which implies that $19^{31}>13^{33}$.
Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?
answered 15 hours ago
Robert ZRobert Z
99.8k1068140
$endgroup$
Note that by the binomial theorem
$$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
which implies that $19^{31}>13^{33}$.
Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?
answered 15 hours ago
Robert ZRobert Z
99.8k1068140
edited 14 hours ago
answered 15 hours ago
Robert ZRobert Z
99.8k1068140
answered 15 hours ago
Robert ZRobert Z
99.8k1068140
answered 15 hours ago
Robert ZRobert Z
99.8k1068140
99.8k1068140
$begingroup$
Solved bonus below(before seeing it posted).
$endgroup$
– miniparser
14 hours ago
add a comment |
$begingroup$
Solved bonus below(before seeing it posted).
$endgroup$
– miniparser
14 hours ago
$begingroup$
Solved bonus below(before seeing it posted).
$endgroup$
– miniparser
14 hours ago
$begingroup$
Solved bonus below(before seeing it posted).
$endgroup$
– miniparser
14 hours ago
add a comment |
$begingroup$
$$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$
$$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$
Clearly $$2^{15}geq 13cdot 9implies A>B$$
answered 15 hours ago
greedoidgreedoid
46.1k1160117
$endgroup$
add a comment |
$begingroup$
$$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$
$$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$
Clearly $$2^{15}geq 13cdot 9implies A>B$$
answered 15 hours ago
greedoidgreedoid
46.1k1160117
$endgroup$
add a comment |
$begingroup$
$$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$
$$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$
Clearly $$2^{15}geq 13cdot 9implies A>B$$
answered 15 hours ago
greedoidgreedoid
46.1k1160117
$endgroup$
$$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$
$$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$
Clearly $$2^{15}geq 13cdot 9implies A>B$$
answered 15 hours ago
greedoidgreedoid
46.1k1160117
answered 15 hours ago
greedoidgreedoid
46.1k1160117
answered 15 hours ago
greedoidgreedoid
46.1k1160117
answered 15 hours ago
greedoidgreedoid
46.1k1160117
46.1k1160117
add a comment |
add a comment |
$begingroup$
Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
$x ln(19) > (x+2) ln(13)$
and finally
$ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.
answered 15 hours ago
Gâteau-GalloisGâteau-Gallois
410213
$endgroup$
add a comment |
$begingroup$
Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
$x ln(19) > (x+2) ln(13)$
and finally
$ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.
answered 15 hours ago
Gâteau-GalloisGâteau-Gallois
410213
$endgroup$
add a comment |
$begingroup$
Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
$x ln(19) > (x+2) ln(13)$
and finally
$ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.
answered 15 hours ago
Gâteau-GalloisGâteau-Gallois
410213
$endgroup$
Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
$x ln(19) > (x+2) ln(13)$
and finally
$ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.
answered 15 hours ago
Gâteau-GalloisGâteau-Gallois
410213
answered 15 hours ago
Gâteau-GalloisGâteau-Gallois
410213
answered 15 hours ago
Gâteau-GalloisGâteau-Gallois
410213
answered 15 hours ago
Gâteau-GalloisGâteau-Gallois
410213
410213
add a comment |
add a comment |
$begingroup$
By your work
$$frac{19^2}{13^2}>2,$$ which says
$$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!
By this way we can prove that
$$19^{31}>13^{35}.$$
answered 11 hours ago
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
By your work
$$frac{19^2}{13^2}>2,$$ which says
$$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!
By this way we can prove that
$$19^{31}>13^{35}.$$
answered 11 hours ago
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
By your work
$$frac{19^2}{13^2}>2,$$ which says
$$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!
By this way we can prove that
$$19^{31}>13^{35}.$$
answered 11 hours ago
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
By your work
$$frac{19^2}{13^2}>2,$$ which says
$$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!
By this way we can prove that
$$19^{31}>13^{35}.$$
answered 11 hours ago
Michael RozenbergMichael Rozenberg
107k1894198
edited 11 hours ago
answered 11 hours ago
Michael RozenbergMichael Rozenberg
107k1894198
answered 11 hours ago
Michael RozenbergMichael Rozenberg
107k1894198
answered 11 hours ago
Michael RozenbergMichael Rozenberg
107k1894198
107k1894198
add a comment |
add a comment |
$begingroup$
$19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$
$6859gt 3times 2197$
$6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$
$27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$
answered 14 hours ago
miniparserminiparser
7871616
$endgroup$
add a comment |
$begingroup$
$19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$
$6859gt 3times 2197$
$6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$
$27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$
answered 14 hours ago
miniparserminiparser
7871616
$endgroup$
add a comment |
$begingroup$
$19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$
$6859gt 3times 2197$
$6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$
$27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$
answered 14 hours ago
miniparserminiparser
7871616
$endgroup$
$19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$
$6859gt 3times 2197$
$6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$
$27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$
answered 14 hours ago
miniparserminiparser
7871616
answered 14 hours ago
miniparserminiparser
7871616
answered 14 hours ago
miniparserminiparser
7871616
answered 14 hours ago
miniparserminiparser
7871616
7871616
add a comment |
add a comment |
$begingroup$
Well, using what you've got so far:
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.
$19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.
=====
For what it's worth one way I might do it if it were up to me is to note.
$19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.
So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.
answered 10 hours ago
fleabloodfleablood
71.9k22687
$endgroup$
add a comment |
$begingroup$
Well, using what you've got so far:
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.
$19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.
=====
For what it's worth one way I might do it if it were up to me is to note.
$19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.
So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.
answered 10 hours ago
fleabloodfleablood
71.9k22687
$endgroup$
add a comment |
$begingroup$
Well, using what you've got so far:
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.
$19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.
=====
For what it's worth one way I might do it if it were up to me is to note.
$19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.
So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.
answered 10 hours ago
fleabloodfleablood
71.9k22687
$endgroup$
Well, using what you've got so far:
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.
$19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.
=====
For what it's worth one way I might do it if it were up to me is to note.
$19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.
$(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.
So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.
answered 10 hours ago
fleabloodfleablood
71.9k22687
edited 10 hours ago
answered 10 hours ago
fleabloodfleablood
71.9k22687
answered 10 hours ago
fleabloodfleablood
71.9k22687
answered 10 hours ago
fleabloodfleablood
71.9k22687
71.9k22687
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add a comment |
$begingroup$
$left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$
answered 7 hours ago
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
add a comment |
$begingroup$
$left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$
answered 7 hours ago
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
add a comment |
$begingroup$
$left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$
answered 7 hours ago
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
$left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$
answered 7 hours ago
Bill DubuqueBill Dubuque
212k29195650
answered 7 hours ago
Bill DubuqueBill Dubuque
212k29195650
answered 7 hours ago
Bill DubuqueBill Dubuque
212k29195650
answered 7 hours ago
Bill DubuqueBill Dubuque
212k29195650
212k29195650
add a comment |
add a comment |
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