Find all roots for the equation [duplicate]

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  Solve for $x$: $2^x=4x$
  
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I am trying to solve $2^x = 4x$. Have taken logs on both sides, represented as an exponent and haven't got it close to the form from which I could find a solution.

algebra-precalculus

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asked Aug 3 at 20:13

Alexander

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marked as duplicate by Dietrich Burde, amWhy algebra-precalculus
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Aug 3 at 20:17

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  See here, and this question.
  – Dietrich Burde
  Aug 3 at 20:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The difference function $xto 2^x-4x$ is convex, there are two solutions, the one is $4$, the other one is numerically $0.30990693238069053545461578388772986095dots$ (as delivered by pari/gp via the command solve( x=0, 3.5, 2^x-4*x )).
  – dan_fulea
  Aug 3 at 20:19
  

  
  
  
  

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This question already has an answer here:

• 
  Solve for $x$: $2^x=4x$
  
  2 answers
  
  

I am trying to solve $2^x = 4x$. Have taken logs on both sides, represented as an exponent and haven't got it close to the form from which I could find a solution.

algebra-precalculus

share|cite|improve this question

asked Aug 3 at 20:13

Alexander

1

marked as duplicate by Dietrich Burde, amWhy algebra-precalculus
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Aug 3 at 20:17

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  See here, and this question.
  – Dietrich Burde
  Aug 3 at 20:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The difference function $xto 2^x-4x$ is convex, there are two solutions, the one is $4$, the other one is numerically $0.30990693238069053545461578388772986095dots$ (as delivered by pari/gp via the command solve( x=0, 3.5, 2^x-4*x )).
  – dan_fulea
  Aug 3 at 20:19
  

  
  
  
  

add a comment | 

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up vote
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favorite

This question already has an answer here:

• 
  Solve for $x$: $2^x=4x$
  
  2 answers
  
  

I am trying to solve $2^x = 4x$. Have taken logs on both sides, represented as an exponent and haven't got it close to the form from which I could find a solution.

algebra-precalculus

share|cite|improve this question

asked Aug 3 at 20:13

Alexander

1

This question already has an answer here:

• 
  Solve for $x$: $2^x=4x$
  
  2 answers
  
  

I am trying to solve $2^x = 4x$. Have taken logs on both sides, represented as an exponent and haven't got it close to the form from which I could find a solution.

This question already has an answer here:

• 
  Solve for $x$: $2^x=4x$
  
  2 answers
  
  

algebra-precalculus

share|cite|improve this question

asked Aug 3 at 20:13

Alexander

1

share|cite|improve this question

share|cite|improve this question

asked Aug 3 at 20:13

Alexander

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asked Aug 3 at 20:13

Alexander

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asked Aug 3 at 20:13

Alexander

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1

marked as duplicate by Dietrich Burde, amWhy algebra-precalculus
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Aug 3 at 20:17

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 Dietrich Burde, amWhy algebra-precalculus
Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

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Aug 3 at 20:17

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  See here, and this question.
  – Dietrich Burde
  Aug 3 at 20:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The difference function $xto 2^x-4x$ is convex, there are two solutions, the one is $4$, the other one is numerically $0.30990693238069053545461578388772986095dots$ (as delivered by pari/gp via the command solve( x=0, 3.5, 2^x-4*x )).
  – dan_fulea
  Aug 3 at 20:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  See here, and this question.
  – Dietrich Burde
  Aug 3 at 20:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The difference function $xto 2^x-4x$ is convex, there are two solutions, the one is $4$, the other one is numerically $0.30990693238069053545461578388772986095dots$ (as delivered by pari/gp via the command solve( x=0, 3.5, 2^x-4*x )).
  – dan_fulea
  Aug 3 at 20:19
  

  
  
  
  

See here, and this question.
– Dietrich Burde
Aug 3 at 20:15

See here, and this question.
– Dietrich Burde
Aug 3 at 20:15

The difference function $xto 2^x-4x$ is convex, there are two solutions, the one is $4$, the other one is numerically $0.30990693238069053545461578388772986095dots$ (as delivered by pari/gp via the command solve( x=0, 3.5, 2^x-4*x )).
– dan_fulea
Aug 3 at 20:19

The difference function $xto 2^x-4x$ is convex, there are two solutions, the one is $4$, the other one is numerically $0.30990693238069053545461578388772986095dots$ (as delivered by pari/gp via the command solve( x=0, 3.5, 2^x-4*x )).
– dan_fulea
Aug 3 at 20:19

add a comment | 

1 Answer
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Note you can write $4=2^2$ so that you have $2^x=2^2x$, dividing by $2^2$ you have $2^x-2=x$. But this is really as far as this goes since these type of equations require special functions to solve. WolframAlpha gives
$$
x= - dfracW_nleft(-dfraclog 24right)log 2
$$
Note that you can try a few 'nice' solutions and see that $x=4$ works.

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answered Aug 3 at 20:17

mathematics2x2life

7,58121636

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1 Answer
1

active

oldest

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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Note you can write $4=2^2$ so that you have $2^x=2^2x$, dividing by $2^2$ you have $2^x-2=x$. But this is really as far as this goes since these type of equations require special functions to solve. WolframAlpha gives
$$
x= - dfracW_nleft(-dfraclog 24right)log 2
$$
Note that you can try a few 'nice' solutions and see that $x=4$ works.

share|cite|improve this answer

answered Aug 3 at 20:17

mathematics2x2life

7,58121636

add a comment | 

up vote
0
down vote

Note you can write $4=2^2$ so that you have $2^x=2^2x$, dividing by $2^2$ you have $2^x-2=x$. But this is really as far as this goes since these type of equations require special functions to solve. WolframAlpha gives
$$
x= - dfracW_nleft(-dfraclog 24right)log 2
$$
Note that you can try a few 'nice' solutions and see that $x=4$ works.

share|cite|improve this answer

answered Aug 3 at 20:17

mathematics2x2life

7,58121636

add a comment | 

up vote
0
down vote

up vote
0
down vote

Note you can write $4=2^2$ so that you have $2^x=2^2x$, dividing by $2^2$ you have $2^x-2=x$. But this is really as far as this goes since these type of equations require special functions to solve. WolframAlpha gives
$$
x= - dfracW_nleft(-dfraclog 24right)log 2
$$
Note that you can try a few 'nice' solutions and see that $x=4$ works.

share|cite|improve this answer

answered Aug 3 at 20:17

mathematics2x2life

7,58121636

Note you can write $4=2^2$ so that you have $2^x=2^2x$, dividing by $2^2$ you have $2^x-2=x$. But this is really as far as this goes since these type of equations require special functions to solve. WolframAlpha gives
$$
x= - dfracW_nleft(-dfraclog 24right)log 2
$$
Note that you can try a few 'nice' solutions and see that $x=4$ works.

share|cite|improve this answer

answered Aug 3 at 20:17

mathematics2x2life

7,58121636

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 20:17

mathematics2x2life

7,58121636

answered Aug 3 at 20:17

mathematics2x2life

7,58121636

answered Aug 3 at 20:17

mathematics2x2life

7,58121636

7,58121636

add a comment | 

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