Solving ln(1+e^-x) = -x+2 for x

[math007]

Homework Statement

I need to solve x:
ln (1+e^-x)=-x+2

Homework Equations

The Attempt at a Solution

ln (1+e^-x)=-x+2
x+ln (1+1/e^x)=2
x+ln (e^x/e^x+1/e^x)=2
x+ln ((e^x+1)/e^x)=2
x+ln (e^x+1)-ln(e^x)=2
x+ln (e^x+1)-x=2
ln (e^x+1)=2im stuck here.
thank you

 

[BiGyElLoWhAt]

Perhaps try raising e to both sides of the equation? $e^{lhs}=e^{rhs}$

 

[math007]

Perhaps try raising e to both sides of the equation? $e^{lhs}=e^{rhs}$

sorry I don't really understand.

 

[BiGyElLoWhAt]

ln (e^x+1)=2...
$e^{ln(e^x+1)}=e^2$
...

 

[BiGyElLoWhAt]

In my previous post, lhs was left hand side and rhs was right hand side. That'll come up from time to time, so it wouldn't hurt to keep that in mind.

 

[math007]

ln (e^x+1)=2...
$e^{ln(e^x+1)}=e^2$
...

where can I read about the rules/ methods to solve ln /e equation because this is confusing me I don't know what to do. :'(

 

[Chestermiller]

where can I read about the rules/ methods to solve ln /e equation because this is confusing me I don't know what to do. :'(

How would you define the natural log of say a variable called y?

 

[BiGyElLoWhAt]

To add to chet, define it in words, not maths, but of course using some math termonology, such as exponents and whatnot.

 

[math007]

How would you define the natural log of say a variable called y?

ln (y)?

 

[BiGyElLoWhAt]

yes
define it in words.

 

[math007]

yes
define it in words.

the exponential form of y?

 

[BiGyElLoWhAt]

What is the natural log of y? If I punch $ln(1)$ in my calculator, it returns a 0, why is the answer 0? what does the 0 represent?

 

[Chestermiller]

Fill in the blanks: The natural log of y is the power to which you have to raise ____ to get ___.

Chet

 

[math007]

to raise e to get y

ln (e^y)=yln e=y

 

[BiGyElLoWhAt]

Ahhh, so ln(y) is the exponent that I can put on e to get y? So what happens if I take that exponent and stick it on e? what do I get? $e^{ln(y)}=?$

 

[Chestermiller]

to raise e to get y

ln (e^y)=yln e=y

Good. Now going back to post #4, fill in the blanks: $\ln(1+e^x)$ is the power to which you have to raise ____ to get _____.

Chet

 

[math007]

Good. Now going back to post #4, fill in the blanks: $\ln(1+e^x)$ is the power to which you have to raise ____ to get _____.

Chet

Good. Now going back to post #4, fill in the blanks: $\ln(1+e^x)$ is the power to which you have to raise ____ to get _____.

Chet

raise e to get ln (1+e^x)
e^(ln (1+e^x))
then e and ln can cancels? to get 1+e^x

 

[Chestermiller]

OK. Good. Now, if you combine this result with the equation in post #4, what do you get for x?

Chet

 

[math007]

OK. Good. Now, if you combine this result with the equation in post #4, what do you get for x?

Chet

raise e to get ln (1+e^x)
e^(ln (1+e^x))
then e and ln can cancels? to get 1+e^x

e^x+1=e^2
e^x=e^2-1
lne^x=ln (e^2-1)
x=ln (e^2-1)

 

[Chestermiller]

e^x+1=e^2
e^x=e^2-1
lne^x=ln (e^2-1)
x=ln (e^2-1)

Excellent!

 

[math007]

Excellent!

awesome thank you so much for the help :-)

 

[Chestermiller]

awesome thank you so much for the help :-)

Most of the thanks should go to Bigyellowwhat.

 

[math007]

Ahhh, so ln(y) is the exponent that I can put on e to get y? So what happens if I take that exponent and stick it on e? what do I get? $e^{ln(y)}=?$

thanks so much for the help man. I really appreciate it :-)

 

[BiGyElLoWhAt]

Anytime