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All real number for which $n$ in $5^n+7^n+11^n=6^n+8^n+9^n$ [duplicate]
This question already has an answer here:
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
4 answers
Finding all real number $n$ in
$$5^n+7^n+11^n=6^n+8^n+9^n$$
Try: From given equation
$n=0,1$ are the solution
But i did not understand any other solution exists or not
Although i have tried like this way
$$bigg(frac{5}{9}bigg)^n+bigg(frac{7}{9}bigg)^n+bigg(frac{11}{9}bigg)^n = bigg(frac{6}{9}bigg)^n+bigg(frac{8}{9}bigg)^n+1$$
Right side is strictly increasing function. but i have a confusion whether left side is strictly increasing or not
could some help me how to solve it, thanks
calculus inequality exponential-sum
edited Nov 21 '18 at 10:09
Batominovski
33.9k33292
asked Nov 21 '18 at 8:26
D Tiwari
5,4132630
marked as duplicate by Michael Rozenberg calculus
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Nov 21 '18 at 12:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
4 answers
Finding all real number $n$ in
$$5^n+7^n+11^n=6^n+8^n+9^n$$
Try: From given equation
$n=0,1$ are the solution
But i did not understand any other solution exists or not
Although i have tried like this way
$$bigg(frac{5}{9}bigg)^n+bigg(frac{7}{9}bigg)^n+bigg(frac{11}{9}bigg)^n = bigg(frac{6}{9}bigg)^n+bigg(frac{8}{9}bigg)^n+1$$
Right side is strictly increasing function. but i have a confusion whether left side is strictly increasing or not
could some help me how to solve it, thanks
calculus inequality exponential-sum
edited Nov 21 '18 at 10:09
Batominovski
33.9k33292
asked Nov 21 '18 at 8:26
D Tiwari
5,4132630
marked as duplicate by Michael Rozenberg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Nov 21 '18 at 12:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
It is not correct that the right hand side of your equation is strictly increasing in $n$. Hint: can you say something about the maximum size of the right hand side?
– ToucanNapoleon
Nov 21 '18 at 8:40
See also here: math.stackexchange.com/questions/2840394
– Michael Rozenberg
Nov 21 '18 at 12:09
add a comment |
This question already has an answer here:
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
4 answers
Finding all real number $n$ in
$$5^n+7^n+11^n=6^n+8^n+9^n$$
Try: From given equation
$n=0,1$ are the solution
But i did not understand any other solution exists or not
Although i have tried like this way
$$bigg(frac{5}{9}bigg)^n+bigg(frac{7}{9}bigg)^n+bigg(frac{11}{9}bigg)^n = bigg(frac{6}{9}bigg)^n+bigg(frac{8}{9}bigg)^n+1$$
Right side is strictly increasing function. but i have a confusion whether left side is strictly increasing or not
could some help me how to solve it, thanks
calculus inequality exponential-sum
edited Nov 21 '18 at 10:09
Batominovski
33.9k33292
asked Nov 21 '18 at 8:26
D Tiwari
5,4132630
This question already has an answer here:
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
4 answers
Finding all real number $n$ in
$$5^n+7^n+11^n=6^n+8^n+9^n$$
Try: From given equation
$n=0,1$ are the solution
But i did not understand any other solution exists or not
Although i have tried like this way
$$bigg(frac{5}{9}bigg)^n+bigg(frac{7}{9}bigg)^n+bigg(frac{11}{9}bigg)^n = bigg(frac{6}{9}bigg)^n+bigg(frac{8}{9}bigg)^n+1$$
Right side is strictly increasing function. but i have a confusion whether left side is strictly increasing or not
could some help me how to solve it, thanks
This question already has an answer here:
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
4 answers
calculus inequality exponential-sum
calculus inequality exponential-sum
edited Nov 21 '18 at 10:09
Batominovski
33.9k33292
asked Nov 21 '18 at 8:26
D Tiwari
5,4132630
edited Nov 21 '18 at 10:09
Batominovski
33.9k33292
asked Nov 21 '18 at 8:26
D Tiwari
5,4132630
edited Nov 21 '18 at 10:09
Batominovski
33.9k33292
edited Nov 21 '18 at 10:09
Batominovski
33.9k33292
edited Nov 21 '18 at 10:09
Batominovski
33.9k33292
33.9k33292
asked Nov 21 '18 at 8:26
D Tiwari
5,4132630
asked Nov 21 '18 at 8:26
D Tiwari
5,4132630
asked Nov 21 '18 at 8:26
D Tiwari
5,4132630
5,4132630
marked as duplicate by Michael Rozenberg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Nov 21 '18 at 12:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Michael Rozenberg calculus
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Nov 21 '18 at 12:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
It is not correct that the right hand side of your equation is strictly increasing in $n$. Hint: can you say something about the maximum size of the right hand side?
– ToucanNapoleon
Nov 21 '18 at 8:40
See also here: math.stackexchange.com/questions/2840394
– Michael Rozenberg
Nov 21 '18 at 12:09
add a comment |
2
It is not correct that the right hand side of your equation is strictly increasing in $n$. Hint: can you say something about the maximum size of the right hand side?
– ToucanNapoleon
Nov 21 '18 at 8:40
See also here: math.stackexchange.com/questions/2840394
– Michael Rozenberg
Nov 21 '18 at 12:09
2
2
It is not correct that the right hand side of your equation is strictly increasing in $n$. Hint: can you say something about the maximum size of the right hand side?
– ToucanNapoleon
Nov 21 '18 at 8:40
It is not correct that the right hand side of your equation is strictly increasing in $n$. Hint: can you say something about the maximum size of the right hand side?
– ToucanNapoleon
Nov 21 '18 at 8:40
See also here: math.stackexchange.com/questions/2840394
– Michael Rozenberg
Nov 21 '18 at 12:09
See also here: math.stackexchange.com/questions/2840394
– Michael Rozenberg
Nov 21 '18 at 12:09
add a comment |
1 Answer
1
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oldest
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Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly concave.
Our equation is equivalent to $f(5) + f(7) + f(11) = f(6) + f(8) + f(9)$
In the convex case we have:
$f(6) = f(frac{5+7}{2}) < frac{f(5)+f(7)}{2}$,
$f(9) = f(frac{7+11}{2}) < frac{f(7)+f(11)}{2}$,
and that
$f(8) = f(frac{11+5}{2}) < frac{f(11)+f(5)}{2}$
So $ f(5) + f(7) + f(11) > f(6) + f(8) + f(9)$
In the concave case the same strict inequalities hold, but reversed.
Thus only n=0 or n=1 could be solutions and they both work.
answered Nov 21 '18 at 9:14
Sorin Tirc
1,520113
1
You should include $n<0$ in your strictly convex case, and for the strictly concave case, $0<n<1$. There is no assumption that $n$ is nonnegative.
– Batominovski
Nov 21 '18 at 9:16
I did'nt exclude n < 0 in the first place :D
– Sorin Tirc
Nov 21 '18 at 9:17
Either you excluded $n<0$, or you made a mistake. The strictly concave case is $0<n<1$, not $n<1$.
– Batominovski
Nov 21 '18 at 9:17
Oops good point :D thanks
– Sorin Tirc
Nov 21 '18 at 9:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly concave.
Our equation is equivalent to $f(5) + f(7) + f(11) = f(6) + f(8) + f(9)$
In the convex case we have:
$f(6) = f(frac{5+7}{2}) < frac{f(5)+f(7)}{2}$,
$f(9) = f(frac{7+11}{2}) < frac{f(7)+f(11)}{2}$,
and that
$f(8) = f(frac{11+5}{2}) < frac{f(11)+f(5)}{2}$
So $ f(5) + f(7) + f(11) > f(6) + f(8) + f(9)$
In the concave case the same strict inequalities hold, but reversed.
Thus only n=0 or n=1 could be solutions and they both work.
answered Nov 21 '18 at 9:14
Sorin Tirc
1,520113
1
You should include $n<0$ in your strictly convex case, and for the strictly concave case, $0<n<1$. There is no assumption that $n$ is nonnegative.
– Batominovski
Nov 21 '18 at 9:16
I did'nt exclude n < 0 in the first place :D
– Sorin Tirc
Nov 21 '18 at 9:17
Either you excluded $n<0$, or you made a mistake. The strictly concave case is $0<n<1$, not $n<1$.
– Batominovski
Nov 21 '18 at 9:17
Oops good point :D thanks
– Sorin Tirc
Nov 21 '18 at 9:18
add a comment |
Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly concave.
Our equation is equivalent to $f(5) + f(7) + f(11) = f(6) + f(8) + f(9)$
In the convex case we have:
$f(6) = f(frac{5+7}{2}) < frac{f(5)+f(7)}{2}$,
$f(9) = f(frac{7+11}{2}) < frac{f(7)+f(11)}{2}$,
and that
$f(8) = f(frac{11+5}{2}) < frac{f(11)+f(5)}{2}$
So $ f(5) + f(7) + f(11) > f(6) + f(8) + f(9)$
In the concave case the same strict inequalities hold, but reversed.
Thus only n=0 or n=1 could be solutions and they both work.
answered Nov 21 '18 at 9:14
Sorin Tirc
1,520113
1
You should include $n<0$ in your strictly convex case, and for the strictly concave case, $0<n<1$. There is no assumption that $n$ is nonnegative.
– Batominovski
Nov 21 '18 at 9:16
I did'nt exclude n < 0 in the first place :D
– Sorin Tirc
Nov 21 '18 at 9:17
Either you excluded $n<0$, or you made a mistake. The strictly concave case is $0<n<1$, not $n<1$.
– Batominovski
Nov 21 '18 at 9:17
Oops good point :D thanks
– Sorin Tirc
Nov 21 '18 at 9:18
add a comment |
Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly concave.
Our equation is equivalent to $f(5) + f(7) + f(11) = f(6) + f(8) + f(9)$
In the convex case we have:
$f(6) = f(frac{5+7}{2}) < frac{f(5)+f(7)}{2}$,
$f(9) = f(frac{7+11}{2}) < frac{f(7)+f(11)}{2}$,
and that
$f(8) = f(frac{11+5}{2}) < frac{f(11)+f(5)}{2}$
So $ f(5) + f(7) + f(11) > f(6) + f(8) + f(9)$
In the concave case the same strict inequalities hold, but reversed.
Thus only n=0 or n=1 could be solutions and they both work.
answered Nov 21 '18 at 9:14
Sorin Tirc
1,520113
Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly concave.
Our equation is equivalent to $f(5) + f(7) + f(11) = f(6) + f(8) + f(9)$
In the convex case we have:
$f(6) = f(frac{5+7}{2}) < frac{f(5)+f(7)}{2}$,
$f(9) = f(frac{7+11}{2}) < frac{f(7)+f(11)}{2}$,
and that
$f(8) = f(frac{11+5}{2}) < frac{f(11)+f(5)}{2}$
So $ f(5) + f(7) + f(11) > f(6) + f(8) + f(9)$
In the concave case the same strict inequalities hold, but reversed.
Thus only n=0 or n=1 could be solutions and they both work.
answered Nov 21 '18 at 9:14
Sorin Tirc
1,520113
edited Nov 21 '18 at 9:18
answered Nov 21 '18 at 9:14
Sorin Tirc
1,520113
answered Nov 21 '18 at 9:14
Sorin Tirc
1,520113
answered Nov 21 '18 at 9:14
Sorin Tirc
1,520113
1,520113
1
You should include $n<0$ in your strictly convex case, and for the strictly concave case, $0<n<1$. There is no assumption that $n$ is nonnegative.
– Batominovski
Nov 21 '18 at 9:16
I did'nt exclude n < 0 in the first place :D
– Sorin Tirc
Nov 21 '18 at 9:17
Either you excluded $n<0$, or you made a mistake. The strictly concave case is $0<n<1$, not $n<1$.
– Batominovski
Nov 21 '18 at 9:17
Oops good point :D thanks
– Sorin Tirc
Nov 21 '18 at 9:18
add a comment |
1
You should include $n<0$ in your strictly convex case, and for the strictly concave case, $0<n<1$. There is no assumption that $n$ is nonnegative.
– Batominovski
Nov 21 '18 at 9:16
I did'nt exclude n < 0 in the first place :D
– Sorin Tirc
Nov 21 '18 at 9:17
Either you excluded $n<0$, or you made a mistake. The strictly concave case is $0<n<1$, not $n<1$.
– Batominovski
Nov 21 '18 at 9:17
Oops good point :D thanks
– Sorin Tirc
Nov 21 '18 at 9:18
1
1
You should include $n<0$ in your strictly convex case, and for the strictly concave case, $0<n<1$. There is no assumption that $n$ is nonnegative.
– Batominovski
Nov 21 '18 at 9:16
You should include $n<0$ in your strictly convex case, and for the strictly concave case, $0<n<1$. There is no assumption that $n$ is nonnegative.
– Batominovski
Nov 21 '18 at 9:16
I did'nt exclude n < 0 in the first place :D
– Sorin Tirc
Nov 21 '18 at 9:17
I did'nt exclude n < 0 in the first place :D
– Sorin Tirc
Nov 21 '18 at 9:17
Either you excluded $n<0$, or you made a mistake. The strictly concave case is $0<n<1$, not $n<1$.
– Batominovski
Nov 21 '18 at 9:17
Either you excluded $n<0$, or you made a mistake. The strictly concave case is $0<n<1$, not $n<1$.
– Batominovski
Nov 21 '18 at 9:17
Oops good point :D thanks
– Sorin Tirc
Nov 21 '18 at 9:18
Oops good point :D thanks
– Sorin Tirc
Nov 21 '18 at 9:18
add a comment |
2
It is not correct that the right hand side of your equation is strictly increasing in $n$. Hint: can you say something about the maximum size of the right hand side?
– ToucanNapoleon
Nov 21 '18 at 8:40
See also here: math.stackexchange.com/questions/2840394
– Michael Rozenberg
Nov 21 '18 at 12:09