Solve Math Log Problems: Prove x+ln(1+e^(-x)) = ln(1+e^(x))

[masterchiefo]

Homework Statement

1. I have to prove that x+ln(1+e^(-x)) is equal to ln(1+e^(x))

Homework Equations

The Attempt at a Solution

1. x+ln(1+e^(-x)) =ln(1+e^(x))
ln(e^(-x)+1)+x
ln(1/(e^x)+1)+x

Im blocked here.. :(

thanks

 

[PeroK]

Hint:

If $e^a = e^b$ what can you say about $a$ and $b$?

 

[masterchiefo]

Hint:

If $e^a = e^b$ what can you say about $a$ and $b$?

that a and b is equal, same number

 

[PeroK]

that a and b is equal, same number

Can you use that fact to help solve your problem?

 

[Mark44]

Homework Statement

1. I have to prove that x+ln(1+e^(-x)) is equal to ln(1+e^(x))

Homework Equations

The Attempt at a Solution

1. x+ln(1+e^(-x)) =ln(1+e^(x))

This is not the best way to start out, since this is what you're supposed to prove.
A better way, IMO, is to start with this:
$x + ln(1 + e^{-x}) - ln(1 + e^x)$
If you can show that this is equal to zero, then you will have proved that the equation is an identity.

Hint: Write the first term as the ln of something, then combine all three terms using the properties of logs.

ln(e^(-x)+1)+x
ln(1/(e^x)+1)+x

Im blocked here.. :(

thanks

 

[HallsofIvy]

If $y= x+ ln(1+ e^{-x})$ then $e^y= e^x(1+ e^{-x})$

If $z= ln(1+ e^x)$ then $e^z= ??$.

 

[Chestermiller]

$$x=\ln (e^x)$$

Chet

 

[masterchiefo]

Sorry guys I still don't understand how to do this, been reading online tutorials on how to solve ln equation with no luck.

 

[ehild]

You need to know that $e^{-x}=\frac{1}{e^x}$ and $\ln(\frac{a}{b})=\ln(a)-\ln(b)$.

write 1+e^-x in fraction form.

 

[masterchiefo]

so 1 + e^(-x) becomes 1 + 1/e^x which we can make one fraction with a common denominator of e^x. so it'll be (e^x/e^x) + (1/e^x)
or (e^x + 1)/(e^x). we can use props of lns to change that into ln(e^x + 1) - ln(e^x)
so it'd be x + ln(e^x + 1) - ln(e^x) = ln(1 + e^x)
for ln(e^x) the ln and e cancel each other, so only the x stays
so it'd be x + ln(e^x + 1) - x = ln(e^x + 1)
and the x and -x cancel each other. so you have ln(e^x+1) = ln (e^x+1)

is this right ?

 

[masterchiefo]

You need to know that $e^{-x}=\frac{1}{e^x}$ and $\ln(\frac{a}{b})=\ln(a)-\ln(b)$.

write 1+e^-x in fraction form.

posted an attempt in my last post

 

[ehild]

so 1 + e^(-x) becomes 1 + 1/e^x which we can make one fraction with a common denominator of e^x. so it'll be (e^x/e^x) + (1/e^x)
or (e^x + 1)/(e^x). we can use props of lns to change that into ln(e^x + 1) - ln(e^x)
so it'd be x + ln(e^x + 1) - ln(e^x) = ln(1 + e^x)
for ln(e^x) the ln and e cancel each other, so only the x stays
so it'd be x + ln(e^x + 1) - x = ln(e^x + 1)
and the x and -x cancel each other. so you have ln(e^x+1) = ln (e^x+1)

is this right ?

It is right.

 

[masterchiefo]

It is right.

awesome man thank you very much.

 

[Chestermiller]

Sorry guys I still don't understand how to do this, been reading online tutorials on how to solve ln equation with no luck.

awesome man thank you very much.

I guess you saw no value in paying attention to the equation I wrote in post # 7. If you substitute this into your original equation, you get:
$$x+\ln (1+e^{-x})=\ln e^x+\ln (1+e^{-x})=\ln [e^x(1+e^{-x})]$$
What does that give you?

Chet

 

[masterchiefo]

I guess you saw no value in paying attention to the equation I wrote in post # 7. If you substitute this into your original equation, you get:
$$x+\ln (1+e^{-x})=\ln e^x+\ln (1+e^{-x})=\ln [e^x(1+e^{-x})]$$
What does that give you?

Chet

yeah sorry I did not really understand what your post meant :/
I am not very good with ln /e/log solver.

 

[Chestermiller]

yeah sorry I did not really understand what your post meant :/
I am not very good with ln /e/log solver.

The definition of lny is the power to which you have to raise e to get y. Now let's substitute e^x for y in the previous statement:

$\ln (e^x)$ is the power to which you have to raise e to get $e^x$. But that's just x.

Chet