Mixing
Any neat proof that $0$ is the unique solution of the equation $4^x+9^x+25^x=6^x+10^x+15^x$?
$begingroup$
It is obvious that both $f(x)= 4^x+9^x+25^x$ and $g(x)=6^x+10^x+15^x$ are strict monotonic increasing functions. It is also easy to check that 0 is a solution of the equation. Also I chart the functions, and it looks that for any $x$, $f(x)>g(x)$, which can be somehow proof by studying the derivative of the $h(x)=f(x)-g(x)$ and showing that $(0,0)$ it's an absolute minimum point for $h(x)$. However $h(x)$ it is a function with a messy derivative, and is not looking easy(for me) to find the zeroes of this derivative.
Does anyone, know an elegant proof(maybe an elementary one, without derivatives) for this problem?
real-analysis exponential-function
asked Aug 18 '17 at 19:18
motorasmotoras
417312
$endgroup$
add a comment |
$begingroup$
It is obvious that both $f(x)= 4^x+9^x+25^x$ and $g(x)=6^x+10^x+15^x$ are strict monotonic increasing functions. It is also easy to check that 0 is a solution of the equation. Also I chart the functions, and it looks that for any $x$, $f(x)>g(x)$, which can be somehow proof by studying the derivative of the $h(x)=f(x)-g(x)$ and showing that $(0,0)$ it's an absolute minimum point for $h(x)$. However $h(x)$ it is a function with a messy derivative, and is not looking easy(for me) to find the zeroes of this derivative.
Does anyone, know an elegant proof(maybe an elementary one, without derivatives) for this problem?
real-analysis exponential-function
asked Aug 18 '17 at 19:18
motorasmotoras
417312
$endgroup$
add a comment |
$begingroup$
It is obvious that both $f(x)= 4^x+9^x+25^x$ and $g(x)=6^x+10^x+15^x$ are strict monotonic increasing functions. It is also easy to check that 0 is a solution of the equation. Also I chart the functions, and it looks that for any $x$, $f(x)>g(x)$, which can be somehow proof by studying the derivative of the $h(x)=f(x)-g(x)$ and showing that $(0,0)$ it's an absolute minimum point for $h(x)$. However $h(x)$ it is a function with a messy derivative, and is not looking easy(for me) to find the zeroes of this derivative.
Does anyone, know an elegant proof(maybe an elementary one, without derivatives) for this problem?
real-analysis exponential-function
asked Aug 18 '17 at 19:18
motorasmotoras
417312
$endgroup$
It is obvious that both $f(x)= 4^x+9^x+25^x$ and $g(x)=6^x+10^x+15^x$ are strict monotonic increasing functions. It is also easy to check that 0 is a solution of the equation. Also I chart the functions, and it looks that for any $x$, $f(x)>g(x)$, which can be somehow proof by studying the derivative of the $h(x)=f(x)-g(x)$ and showing that $(0,0)$ it's an absolute minimum point for $h(x)$. However $h(x)$ it is a function with a messy derivative, and is not looking easy(for me) to find the zeroes of this derivative.
Does anyone, know an elegant proof(maybe an elementary one, without derivatives) for this problem?
real-analysis exponential-function
real-analysis exponential-function
asked Aug 18 '17 at 19:18
motorasmotoras
417312
asked Aug 18 '17 at 19:18
motorasmotoras
417312
asked Aug 18 '17 at 19:18
motorasmotoras
417312
asked Aug 18 '17 at 19:18
motorasmotoras
417312
asked Aug 18 '17 at 19:18
motorasmotoras
417312
417312
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
HINT: $$a^2+b^2+c^2geq ab+bc+ca$$
answered Aug 18 '17 at 19:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
Thank you very much, I got it!
$endgroup$
– motoras
Aug 18 '17 at 19:22
1
$begingroup$
ok, that is nice!
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:23
$begingroup$
Very nice observation!
$endgroup$
– Cauchy
Aug 18 '17 at 19:23
2
$begingroup$
From $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
$endgroup$
– Mark Bennet
Aug 18 '17 at 19:25
$begingroup$
at the University i gave a talk about "solving equations with inequalities"
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:25
|
show 2 more comments
$begingroup$
Since the question is already settled for the continuous version, I thought I would just add (for fun!) a short proof for the case in which $x$ is a whole number: for $x geq 1$, the two sides are not equal when reduced modulo ten. So, the only possible whole number solution is at $x=0$, which works. "QED"
answered Aug 18 '17 at 19:55
Benjamin DickmanBenjamin Dickman
10.3k22968
$endgroup$
add a comment |
$begingroup$
I guess you want a proof that works for the continuous case. If we would know that $x$ is always an integer, then we could say that $f(x)$ is an even number for any $x$ integer except $0$, and $g(x)$ is odd for any $x$ integer except $0$. So, $x$ must be zero.
edited Jan 20 at 13:39
Thomas Shelby
3,7492525
answered Apr 14 '18 at 21:05
FerrisFerris
8815
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT: $$a^2+b^2+c^2geq ab+bc+ca$$
answered Aug 18 '17 at 19:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
Thank you very much, I got it!
$endgroup$
– motoras
Aug 18 '17 at 19:22
1
$begingroup$
ok, that is nice!
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:23
$begingroup$
Very nice observation!
$endgroup$
– Cauchy
Aug 18 '17 at 19:23
2
$begingroup$
From $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
$endgroup$
– Mark Bennet
Aug 18 '17 at 19:25
$begingroup$
at the University i gave a talk about "solving equations with inequalities"
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:25
|
show 2 more comments
$begingroup$
HINT: $$a^2+b^2+c^2geq ab+bc+ca$$
answered Aug 18 '17 at 19:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
Thank you very much, I got it!
$endgroup$
– motoras
Aug 18 '17 at 19:22
1
$begingroup$
ok, that is nice!
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:23
$begingroup$
Very nice observation!
$endgroup$
– Cauchy
Aug 18 '17 at 19:23
2
$begingroup$
From $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
$endgroup$
– Mark Bennet
Aug 18 '17 at 19:25
$begingroup$
at the University i gave a talk about "solving equations with inequalities"
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:25
|
show 2 more comments
$begingroup$
HINT: $$a^2+b^2+c^2geq ab+bc+ca$$
answered Aug 18 '17 at 19:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
HINT: $$a^2+b^2+c^2geq ab+bc+ca$$
answered Aug 18 '17 at 19:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Aug 18 '17 at 19:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Aug 18 '17 at 19:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Aug 18 '17 at 19:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
76.8k42866
$begingroup$
Thank you very much, I got it!
$endgroup$
– motoras
Aug 18 '17 at 19:22
1
$begingroup$
ok, that is nice!
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:23
$begingroup$
Very nice observation!
$endgroup$
– Cauchy
Aug 18 '17 at 19:23
2
$begingroup$
From $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
$endgroup$
– Mark Bennet
Aug 18 '17 at 19:25
$begingroup$
at the University i gave a talk about "solving equations with inequalities"
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:25
|
show 2 more comments
$begingroup$
Thank you very much, I got it!
$endgroup$
– motoras
Aug 18 '17 at 19:22
1
$begingroup$
ok, that is nice!
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:23
$begingroup$
Very nice observation!
$endgroup$
– Cauchy
Aug 18 '17 at 19:23
2
$begingroup$
From $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
$endgroup$
– Mark Bennet
Aug 18 '17 at 19:25
$begingroup$
at the University i gave a talk about "solving equations with inequalities"
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:25
$begingroup$
Thank you very much, I got it!
$endgroup$
– motoras
Aug 18 '17 at 19:22
$begingroup$
Thank you very much, I got it!
$endgroup$
– motoras
Aug 18 '17 at 19:22
1
1
$begingroup$
ok, that is nice!
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:23
$begingroup$
ok, that is nice!
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:23
$begingroup$
Very nice observation!
$endgroup$
– Cauchy
Aug 18 '17 at 19:23
$begingroup$
Very nice observation!
$endgroup$
– Cauchy
Aug 18 '17 at 19:23
2
2
$begingroup$
From $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
$endgroup$
– Mark Bennet
Aug 18 '17 at 19:25
$begingroup$
From $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
$endgroup$
– Mark Bennet
Aug 18 '17 at 19:25
$begingroup$
at the University i gave a talk about "solving equations with inequalities"
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:25
$begingroup$
at the University i gave a talk about "solving equations with inequalities"
$endgroup$
– Dr. Sonnhard Graubner
Aug 18 '17 at 19:25
|
show 2 more comments
$begingroup$
Since the question is already settled for the continuous version, I thought I would just add (for fun!) a short proof for the case in which $x$ is a whole number: for $x geq 1$, the two sides are not equal when reduced modulo ten. So, the only possible whole number solution is at $x=0$, which works. "QED"
answered Aug 18 '17 at 19:55
Benjamin DickmanBenjamin Dickman
10.3k22968
$endgroup$
add a comment |
$begingroup$
Since the question is already settled for the continuous version, I thought I would just add (for fun!) a short proof for the case in which $x$ is a whole number: for $x geq 1$, the two sides are not equal when reduced modulo ten. So, the only possible whole number solution is at $x=0$, which works. "QED"
answered Aug 18 '17 at 19:55
Benjamin DickmanBenjamin Dickman
10.3k22968
$endgroup$
add a comment |
$begingroup$
Since the question is already settled for the continuous version, I thought I would just add (for fun!) a short proof for the case in which $x$ is a whole number: for $x geq 1$, the two sides are not equal when reduced modulo ten. So, the only possible whole number solution is at $x=0$, which works. "QED"
answered Aug 18 '17 at 19:55
Benjamin DickmanBenjamin Dickman
10.3k22968
$endgroup$
Since the question is already settled for the continuous version, I thought I would just add (for fun!) a short proof for the case in which $x$ is a whole number: for $x geq 1$, the two sides are not equal when reduced modulo ten. So, the only possible whole number solution is at $x=0$, which works. "QED"
answered Aug 18 '17 at 19:55
Benjamin DickmanBenjamin Dickman
10.3k22968
edited Aug 19 '17 at 0:19
answered Aug 18 '17 at 19:55
Benjamin DickmanBenjamin Dickman
10.3k22968
answered Aug 18 '17 at 19:55
Benjamin DickmanBenjamin Dickman
10.3k22968
answered Aug 18 '17 at 19:55
Benjamin DickmanBenjamin Dickman
10.3k22968
10.3k22968
add a comment |
add a comment |
$begingroup$
I guess you want a proof that works for the continuous case. If we would know that $x$ is always an integer, then we could say that $f(x)$ is an even number for any $x$ integer except $0$, and $g(x)$ is odd for any $x$ integer except $0$. So, $x$ must be zero.
edited Jan 20 at 13:39
Thomas Shelby
3,7492525
answered Apr 14 '18 at 21:05
FerrisFerris
8815
$endgroup$
add a comment |
$begingroup$
I guess you want a proof that works for the continuous case. If we would know that $x$ is always an integer, then we could say that $f(x)$ is an even number for any $x$ integer except $0$, and $g(x)$ is odd for any $x$ integer except $0$. So, $x$ must be zero.
edited Jan 20 at 13:39
Thomas Shelby
3,7492525
answered Apr 14 '18 at 21:05
FerrisFerris
8815
$endgroup$
add a comment |
$begingroup$
I guess you want a proof that works for the continuous case. If we would know that $x$ is always an integer, then we could say that $f(x)$ is an even number for any $x$ integer except $0$, and $g(x)$ is odd for any $x$ integer except $0$. So, $x$ must be zero.
edited Jan 20 at 13:39
Thomas Shelby
3,7492525
answered Apr 14 '18 at 21:05
FerrisFerris
8815
$endgroup$
I guess you want a proof that works for the continuous case. If we would know that $x$ is always an integer, then we could say that $f(x)$ is an even number for any $x$ integer except $0$, and $g(x)$ is odd for any $x$ integer except $0$. So, $x$ must be zero.
edited Jan 20 at 13:39
Thomas Shelby
3,7492525
answered Apr 14 '18 at 21:05
FerrisFerris
8815
edited Jan 20 at 13:39
Thomas Shelby
3,7492525
edited Jan 20 at 13:39
Thomas Shelby
3,7492525
edited Jan 20 at 13:39
Thomas Shelby
3,7492525
3,7492525
answered Apr 14 '18 at 21:05
FerrisFerris
8815
answered Apr 14 '18 at 21:05
FerrisFerris
8815
answered Apr 14 '18 at 21:05
FerrisFerris
8815
8815
add a comment |
add a comment |
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