Roots of $f(x)=3^x+4^x-5^x$ [duplicate] The Next CEO of Stack OverflowProving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $Solving $1^x+2^x+3^x=0$ equations…Set of limit points of continuous functionsReal Analysis Monotone Convergence Theorem QuestionQuestion about disproving if $exists x_0 : f(x_0)=g(x_0)$ then $exists x_1 : f(x_1)>g(x_1)$If $f$ is nonnegative and measurable then its integral is the limit of integrals of truncated functions$epsilon$ - $delta$ definition of a limit - smaller $epsilon$ implies smaller $delta$?Continuity of an increasing $f$ in terms of sequencesIf $f$ is increasing and continuous at $x$, then there exists two sequences$ a_n $ and $ b_n $, $a_n < x < b_n$Continuous extension simple questionClosure and Sequence
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Roots of $f(x)=3^x+4^x-5^x$ [duplicate]
The Next CEO of Stack OverflowProving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $Solving $1^x+2^x+3^x=0$ equations…Set of limit points of continuous functionsReal Analysis Monotone Convergence Theorem QuestionQuestion about disproving if $exists x_0 : f(x_0)=g(x_0)$ then $exists x_1 : f(x_1)>g(x_1)$If $f$ is nonnegative and measurable then its integral is the limit of integrals of truncated functions$epsilon$ - $delta$ definition of a limit - smaller $epsilon$ implies smaller $delta$?Continuity of an increasing $f$ in terms of sequencesIf $f$ is increasing and continuous at $x$, then there exists two sequences$ a_n $ and $ b_n $, $a_n < x < b_n$Continuous extension simple questionClosure and Sequence
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This question already has an answer here:
Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $
2 answers
Let define $f:mathbbRrightarrowmathbbR$ as $f(x)=3^x+4^x-5^x$.
Prove that there exists only one $x_0$ such as $f(x_0)=0$.
My approach:
We can see that $lim_xrightarrow-inftyf(x)=0$ and $f(2)=0$. I don't know how to use the derivatives of $f$ to prove that it is firstly increasing and then decreasing for $xin(-infty,2)$ and for $xin(2,infty)$ it is only decreasing.
real-analysis roots
asked Mar 20 at 10:02
avan1235avan1235
3578
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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.
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This question already has an answer here:
Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $
2 answers
Let define $f:mathbbRrightarrowmathbbR$ as $f(x)=3^x+4^x-5^x$.
Prove that there exists only one $x_0$ such as $f(x_0)=0$.
My approach:
We can see that $lim_xrightarrow-inftyf(x)=0$ and $f(2)=0$. I don't know how to use the derivatives of $f$ to prove that it is firstly increasing and then decreasing for $xin(-infty,2)$ and for $xin(2,infty)$ it is only decreasing.
real-analysis roots
asked Mar 20 at 10:02
avan1235avan1235
3578
$endgroup$
marked as duplicate by Dietrich Burde, Thomas Shelby, Javi, José Carlos Santos real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Mar 20 at 16:16
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.
1
$begingroup$
math.stackexchange.com/questions/61812/…
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– lab bhattacharjee
Mar 20 at 10:11
$begingroup$
It's a more interesting question when the solutions to this sort of equation are allowed to be complex : math.stackexchange.com/questions/3140553/…
$endgroup$
– Martin Hansen
Mar 20 at 10:45
add a comment |
$begingroup$
This question already has an answer here:
Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $
2 answers
Let define $f:mathbbRrightarrowmathbbR$ as $f(x)=3^x+4^x-5^x$.
Prove that there exists only one $x_0$ such as $f(x_0)=0$.
My approach:
We can see that $lim_xrightarrow-inftyf(x)=0$ and $f(2)=0$. I don't know how to use the derivatives of $f$ to prove that it is firstly increasing and then decreasing for $xin(-infty,2)$ and for $xin(2,infty)$ it is only decreasing.
real-analysis roots
asked Mar 20 at 10:02
avan1235avan1235
3578
$endgroup$
This question already has an answer here:
Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $
2 answers
Let define $f:mathbbRrightarrowmathbbR$ as $f(x)=3^x+4^x-5^x$.
Prove that there exists only one $x_0$ such as $f(x_0)=0$.
My approach:
We can see that $lim_xrightarrow-inftyf(x)=0$ and $f(2)=0$. I don't know how to use the derivatives of $f$ to prove that it is firstly increasing and then decreasing for $xin(-infty,2)$ and for $xin(2,infty)$ it is only decreasing.
This question already has an answer here:
Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $
2 answers
real-analysis roots
real-analysis roots
asked Mar 20 at 10:02
avan1235avan1235
3578
asked Mar 20 at 10:02
avan1235avan1235
3578
asked Mar 20 at 10:02
avan1235avan1235
3578
asked Mar 20 at 10:02
avan1235avan1235
3578
asked Mar 20 at 10:02
avan1235avan1235
3578
3578
marked as duplicate by Dietrich Burde, Thomas Shelby, Javi, José Carlos Santos real-analysis
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marked as duplicate by Dietrich Burde, Thomas Shelby, Javi, José Carlos Santos real-analysis
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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.
1
$begingroup$
math.stackexchange.com/questions/61812/…
$endgroup$
– lab bhattacharjee
Mar 20 at 10:11
$begingroup$
It's a more interesting question when the solutions to this sort of equation are allowed to be complex : math.stackexchange.com/questions/3140553/…
$endgroup$
– Martin Hansen
Mar 20 at 10:45
add a comment |
1
$begingroup$
math.stackexchange.com/questions/61812/…
$endgroup$
– lab bhattacharjee
Mar 20 at 10:11
$begingroup$
It's a more interesting question when the solutions to this sort of equation are allowed to be complex : math.stackexchange.com/questions/3140553/…
$endgroup$
– Martin Hansen
Mar 20 at 10:45
1
1
$begingroup$
math.stackexchange.com/questions/61812/…
$endgroup$
– lab bhattacharjee
Mar 20 at 10:11
$begingroup$
math.stackexchange.com/questions/61812/…
$endgroup$
– lab bhattacharjee
Mar 20 at 10:11
$begingroup$
It's a more interesting question when the solutions to this sort of equation are allowed to be complex : math.stackexchange.com/questions/3140553/…
$endgroup$
– Martin Hansen
Mar 20 at 10:45
$begingroup$
It's a more interesting question when the solutions to this sort of equation are allowed to be complex : math.stackexchange.com/questions/3140553/…
$endgroup$
– Martin Hansen
Mar 20 at 10:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $x>2$ we have $3^x+4^x=(9)3^x-2+(16)4^x-2<(9)5^x-2+(16)5^x-2=5^x$ and the inequalities gets reversed for $x<2$. Hence $x=2$ is the only solution.
answered Mar 20 at 10:10
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
add a comment |
$begingroup$
Easier way to see it is
$$
3^x+4^x=5^x qquadLongleftrightarrowqquad f(x)=left(frac35right)^x
+left(frac45right)^x-1=0
$$
Clearly, the left hand side is a strictly decreasing function as a sum of two strictly decreasing functions.
Also $f(2)=0$, and hence $x=2$ is the one and only root of $f$.
answered Mar 20 at 10:20
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $x>2$ we have $3^x+4^x=(9)3^x-2+(16)4^x-2<(9)5^x-2+(16)5^x-2=5^x$ and the inequalities gets reversed for $x<2$. Hence $x=2$ is the only solution.
answered Mar 20 at 10:10
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
add a comment |
$begingroup$
For $x>2$ we have $3^x+4^x=(9)3^x-2+(16)4^x-2<(9)5^x-2+(16)5^x-2=5^x$ and the inequalities gets reversed for $x<2$. Hence $x=2$ is the only solution.
answered Mar 20 at 10:10
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
add a comment |
$begingroup$
For $x>2$ we have $3^x+4^x=(9)3^x-2+(16)4^x-2<(9)5^x-2+(16)5^x-2=5^x$ and the inequalities gets reversed for $x<2$. Hence $x=2$ is the only solution.
answered Mar 20 at 10:10
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
For $x>2$ we have $3^x+4^x=(9)3^x-2+(16)4^x-2<(9)5^x-2+(16)5^x-2=5^x$ and the inequalities gets reversed for $x<2$. Hence $x=2$ is the only solution.
answered Mar 20 at 10:10
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Mar 20 at 10:10
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Mar 20 at 10:10
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Mar 20 at 10:10
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
71.5k53170
add a comment |
add a comment |
$begingroup$
Easier way to see it is
$$
3^x+4^x=5^x qquadLongleftrightarrowqquad f(x)=left(frac35right)^x
+left(frac45right)^x-1=0
$$
Clearly, the left hand side is a strictly decreasing function as a sum of two strictly decreasing functions.
Also $f(2)=0$, and hence $x=2$ is the one and only root of $f$.
answered Mar 20 at 10:20
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
add a comment |
$begingroup$
Easier way to see it is
$$
3^x+4^x=5^x qquadLongleftrightarrowqquad f(x)=left(frac35right)^x
+left(frac45right)^x-1=0
$$
Clearly, the left hand side is a strictly decreasing function as a sum of two strictly decreasing functions.
Also $f(2)=0$, and hence $x=2$ is the one and only root of $f$.
answered Mar 20 at 10:20
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
add a comment |
$begingroup$
Easier way to see it is
$$
3^x+4^x=5^x qquadLongleftrightarrowqquad f(x)=left(frac35right)^x
+left(frac45right)^x-1=0
$$
Clearly, the left hand side is a strictly decreasing function as a sum of two strictly decreasing functions.
Also $f(2)=0$, and hence $x=2$ is the one and only root of $f$.
answered Mar 20 at 10:20
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
Easier way to see it is
$$
3^x+4^x=5^x qquadLongleftrightarrowqquad f(x)=left(frac35right)^x
+left(frac45right)^x-1=0
$$
Clearly, the left hand side is a strictly decreasing function as a sum of two strictly decreasing functions.
Also $f(2)=0$, and hence $x=2$ is the one and only root of $f$.
answered Mar 20 at 10:20
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Mar 20 at 10:20
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Mar 20 at 10:20
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Mar 20 at 10:20
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
63.7k1385165
add a comment |
add a comment |
1
$begingroup$
math.stackexchange.com/questions/61812/…
$endgroup$
– lab bhattacharjee
Mar 20 at 10:11
$begingroup$
It's a more interesting question when the solutions to this sort of equation are allowed to be complex : math.stackexchange.com/questions/3140553/…
$endgroup$
– Martin Hansen
Mar 20 at 10:45