Solving of the equation $e^-xpi-2e^-2xpi=0$

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0
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If got problems with the answer for $x$ for the equation:

$$e^-xpi-2e^-2xpi=0$$

I don't really know how to start. I tried to substitute $e^x$ but that didn't really work...

I would be happy if someone could provide a little hint for me!
Thanks!

exponential-function substitution exponential-distribution

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edited Jul 22 at 13:49

Rumpelstiltskin

1,409315

asked Jul 22 at 13:31

bikidriver

31

add a comment | 

up vote
0
down vote

favorite

If got problems with the answer for $x$ for the equation:

$$e^-xpi-2e^-2xpi=0$$

I don't really know how to start. I tried to substitute $e^x$ but that didn't really work...

I would be happy if someone could provide a little hint for me!
Thanks!

exponential-function substitution exponential-distribution

share|cite|improve this question

edited Jul 22 at 13:49

Rumpelstiltskin

1,409315

asked Jul 22 at 13:31

bikidriver

31

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

If got problems with the answer for $x$ for the equation:

$$e^-xpi-2e^-2xpi=0$$

I don't really know how to start. I tried to substitute $e^x$ but that didn't really work...

I would be happy if someone could provide a little hint for me!
Thanks!

exponential-function substitution exponential-distribution

share|cite|improve this question

edited Jul 22 at 13:49

Rumpelstiltskin

1,409315

asked Jul 22 at 13:31

bikidriver

31

If got problems with the answer for $x$ for the equation:

$$e^-xpi-2e^-2xpi=0$$

I don't really know how to start. I tried to substitute $e^x$ but that didn't really work...

I would be happy if someone could provide a little hint for me!
Thanks!

exponential-function substitution exponential-distribution

share|cite|improve this question

edited Jul 22 at 13:49

Rumpelstiltskin

1,409315

asked Jul 22 at 13:31

bikidriver

31

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 13:49

Rumpelstiltskin

1,409315

edited Jul 22 at 13:49

Rumpelstiltskin

1,409315

edited Jul 22 at 13:49

Rumpelstiltskin

1,409315

1,409315

asked Jul 22 at 13:31

bikidriver

31

asked Jul 22 at 13:31

bikidriver

31

asked Jul 22 at 13:31

bikidriver

31

31

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

up vote
1
down vote



accepted

Hint: introduce a new variable $y:= e^-pi x$. Also keep in mind that the exponential function is positive everywhere.

share|cite|improve this answer

answered Jul 22 at 13:34

A. Pongrácz

2,109221

• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok lets see: y-2y^2=0 y(1-2y)=0 y1=0 y2=-0,5
  – bikidriver
  Jul 22 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok got it now! thx a lot!
  – bikidriver
  Jul 22 at 13:41
  

  
  
  
  

add a comment | 

up vote
2
down vote

$$e^ -xπ -2e^ -2xπ =0\ e^ -xπ =2e^ -2xπ \ e^ xπ =2\ x=frac ln 2 pi $$

share|cite|improve this answer

answered Jul 22 at 13:37

haqnatural

20.5k72457

add a comment | 

up vote
2
down vote

Come on,

$$e^-xpi=2e^-2xpiiff -xpi=ln2-2xpi.$$

share|cite|improve this answer

answered Jul 22 at 13:37

Yves Daoust

111k665203

add a comment | 

up vote
1
down vote

Let $y=e^-xpi$ so $y=2y^2$. Therefore...

share|cite|improve this answer

answered Jul 22 at 13:34

J.G.

13.2k11424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  got it :) y1=0 ; y2=0,5
  – bikidriver
  Jul 22 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  therefore e^(−xπ)=0,5 --> x=-ln(0,5)/π
  – bikidriver
  Jul 22 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you a lot for the hint!
  – bikidriver
  Jul 22 at 13:45
  

  
  
  
  

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Hint: introduce a new variable $y:= e^-pi x$. Also keep in mind that the exponential function is positive everywhere.

share|cite|improve this answer

answered Jul 22 at 13:34

A. Pongrácz

2,109221

• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok lets see: y-2y^2=0 y(1-2y)=0 y1=0 y2=-0,5
  – bikidriver
  Jul 22 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok got it now! thx a lot!
  – bikidriver
  Jul 22 at 13:41
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Hint: introduce a new variable $y:= e^-pi x$. Also keep in mind that the exponential function is positive everywhere.

share|cite|improve this answer

answered Jul 22 at 13:34

A. Pongrácz

2,109221

• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok lets see: y-2y^2=0 y(1-2y)=0 y1=0 y2=-0,5
  – bikidriver
  Jul 22 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok got it now! thx a lot!
  – bikidriver
  Jul 22 at 13:41
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Hint: introduce a new variable $y:= e^-pi x$. Also keep in mind that the exponential function is positive everywhere.

share|cite|improve this answer

answered Jul 22 at 13:34

A. Pongrácz

2,109221

Hint: introduce a new variable $y:= e^-pi x$. Also keep in mind that the exponential function is positive everywhere.

share|cite|improve this answer

answered Jul 22 at 13:34

A. Pongrácz

2,109221

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 13:34

A. Pongrácz

2,109221

answered Jul 22 at 13:34

A. Pongrácz

2,109221

answered Jul 22 at 13:34

A. Pongrácz

2,109221

2,109221

• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok lets see: y-2y^2=0 y(1-2y)=0 y1=0 y2=-0,5
  – bikidriver
  Jul 22 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok got it now! thx a lot!
  – bikidriver
  Jul 22 at 13:41
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok lets see: y-2y^2=0 y(1-2y)=0 y1=0 y2=-0,5
  – bikidriver
  Jul 22 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok got it now! thx a lot!
  – bikidriver
  Jul 22 at 13:41
  

  
  
  
  

ok lets see: y-2y^2=0 y(1-2y)=0 y1=0 y2=-0,5
– bikidriver
Jul 22 at 13:38

ok lets see: y-2y^2=0 y(1-2y)=0 y1=0 y2=-0,5
– bikidriver
Jul 22 at 13:38

Ok got it now! thx a lot!
– bikidriver
Jul 22 at 13:41

Ok got it now! thx a lot!
– bikidriver
Jul 22 at 13:41

add a comment | 

up vote
2
down vote

$$e^ -xπ -2e^ -2xπ =0\ e^ -xπ =2e^ -2xπ \ e^ xπ =2\ x=frac ln 2 pi $$

share|cite|improve this answer

answered Jul 22 at 13:37

haqnatural

20.5k72457

add a comment | 

up vote
2
down vote

$$e^ -xπ -2e^ -2xπ =0\ e^ -xπ =2e^ -2xπ \ e^ xπ =2\ x=frac ln 2 pi $$

share|cite|improve this answer

answered Jul 22 at 13:37

haqnatural

20.5k72457

add a comment | 

up vote
2
down vote

up vote
2
down vote

$$e^ -xπ -2e^ -2xπ =0\ e^ -xπ =2e^ -2xπ \ e^ xπ =2\ x=frac ln 2 pi $$

share|cite|improve this answer

answered Jul 22 at 13:37

haqnatural

20.5k72457

$$e^ -xπ -2e^ -2xπ =0\ e^ -xπ =2e^ -2xπ \ e^ xπ =2\ x=frac ln 2 pi $$

share|cite|improve this answer

answered Jul 22 at 13:37

haqnatural

20.5k72457

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 13:37

haqnatural

20.5k72457

answered Jul 22 at 13:37

haqnatural

20.5k72457

answered Jul 22 at 13:37

haqnatural

20.5k72457

20.5k72457

add a comment | 

add a comment | 

up vote
2
down vote

Come on,

$$e^-xpi=2e^-2xpiiff -xpi=ln2-2xpi.$$

share|cite|improve this answer

answered Jul 22 at 13:37

Yves Daoust

111k665203

add a comment | 

up vote
2
down vote

Come on,

$$e^-xpi=2e^-2xpiiff -xpi=ln2-2xpi.$$

share|cite|improve this answer

answered Jul 22 at 13:37

Yves Daoust

111k665203

add a comment | 

up vote
2
down vote

up vote
2
down vote

Come on,

$$e^-xpi=2e^-2xpiiff -xpi=ln2-2xpi.$$

share|cite|improve this answer

answered Jul 22 at 13:37

Yves Daoust

111k665203

Come on,

$$e^-xpi=2e^-2xpiiff -xpi=ln2-2xpi.$$

share|cite|improve this answer

answered Jul 22 at 13:37

Yves Daoust

111k665203

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 13:37

Yves Daoust

111k665203

answered Jul 22 at 13:37

Yves Daoust

111k665203

answered Jul 22 at 13:37

Yves Daoust

111k665203

111k665203

add a comment | 

add a comment | 

up vote
1
down vote

Let $y=e^-xpi$ so $y=2y^2$. Therefore...

share|cite|improve this answer

answered Jul 22 at 13:34

J.G.

13.2k11424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  got it :) y1=0 ; y2=0,5
  – bikidriver
  Jul 22 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  therefore e^(−xπ)=0,5 --> x=-ln(0,5)/π
  – bikidriver
  Jul 22 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you a lot for the hint!
  – bikidriver
  Jul 22 at 13:45
  

  
  
  
  

add a comment | 

up vote
1
down vote

Let $y=e^-xpi$ so $y=2y^2$. Therefore...

share|cite|improve this answer

answered Jul 22 at 13:34

J.G.

13.2k11424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  got it :) y1=0 ; y2=0,5
  – bikidriver
  Jul 22 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  therefore e^(−xπ)=0,5 --> x=-ln(0,5)/π
  – bikidriver
  Jul 22 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you a lot for the hint!
  – bikidriver
  Jul 22 at 13:45
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $y=e^-xpi$ so $y=2y^2$. Therefore...

share|cite|improve this answer

answered Jul 22 at 13:34

J.G.

13.2k11424

Let $y=e^-xpi$ so $y=2y^2$. Therefore...

share|cite|improve this answer

answered Jul 22 at 13:34

J.G.

13.2k11424

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 13:34

J.G.

13.2k11424

answered Jul 22 at 13:34

J.G.

13.2k11424

answered Jul 22 at 13:34

J.G.

13.2k11424

13.2k11424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  got it :) y1=0 ; y2=0,5
  – bikidriver
  Jul 22 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  therefore e^(−xπ)=0,5 --> x=-ln(0,5)/π
  – bikidriver
  Jul 22 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you a lot for the hint!
  – bikidriver
  Jul 22 at 13:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  got it :) y1=0 ; y2=0,5
  – bikidriver
  Jul 22 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  therefore e^(−xπ)=0,5 --> x=-ln(0,5)/π
  – bikidriver
  Jul 22 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you a lot for the hint!
  – bikidriver
  Jul 22 at 13:45
  

  
  
  
  

got it :) y1=0 ; y2=0,5
– bikidriver
Jul 22 at 13:43

got it :) y1=0 ; y2=0,5
– bikidriver
Jul 22 at 13:43

therefore e^(−xπ)=0,5 --> x=-ln(0,5)/π
– bikidriver
Jul 22 at 13:44

therefore e^(−xπ)=0,5 --> x=-ln(0,5)/π
– bikidriver
Jul 22 at 13:44

Thank you a lot for the hint!
– bikidriver
Jul 22 at 13:45

Thank you a lot for the hint!
– bikidriver
Jul 22 at 13:45

add a comment | 

 

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