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Roots of $f(x)=3^x+4^x-5^x$ [duplicate]

0

$begingroup$

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

share|cite|improve this question

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/…
  $endgroup$
  – 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 | 

0

$begingroup$

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

share|cite|improve this question

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.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
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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/…
  $endgroup$
  – 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 | 

$begingroup$

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

share|cite|improve this question

asked Mar 20 at 10:02

avan1235avan1235

3578

$endgroup$

share|cite|improve this question

asked Mar 20 at 10:02

avan1235avan1235

3578

share|cite|improve this question

asked Mar 20 at 10:02

avan1235avan1235

3578

asked Mar 20 at 10:02

avan1235avan1235

3578

asked Mar 20 at 10:02

avan1235avan1235

3578

2 Answers
2

active

oldest

votes

4

$begingroup$

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.

share|cite|improve this answer

answered Mar 20 at 10:10

Kavi Rama MurthyKavi Rama Murthy

71.5k53170

$endgroup$

add a comment | 

2

$begingroup$

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 $f$.

share|cite|improve this answer

answered Mar 20 at 10:20

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.7k1385165

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered Mar 20 at 10:10

Kavi Rama MurthyKavi Rama Murthy

71.5k53170

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 20 at 10:10

Kavi Rama MurthyKavi Rama Murthy

71.5k53170

$endgroup$

share|cite|improve this answer

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

2

$begingroup$

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 $f$.

share|cite|improve this answer

answered Mar 20 at 10:20

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.7k1385165

$endgroup$

add a comment | 

2

$begingroup$

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 $f$.

share|cite|improve this answer

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 $f$.

share|cite|improve this answer

answered Mar 20 at 10:20

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.7k1385165

$endgroup$

share|cite|improve this answer

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