# Math Log problems

1. Apr 1, 2015

### masterchiefo

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

2. Apr 1, 2015

### PeroK

Hint:

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

3. Apr 1, 2015

### masterchiefo

that a and b is equal, same number

4. Apr 1, 2015

### PeroK

Can you use that fact to help solve your problem?

5. Apr 1, 2015

### Staff: Mentor

This is not the best way to start out, since this is what you're supposed to prove.
$x + ln(1 + e^{-x}) - ln(1 + e^x)$
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.

6. Apr 2, 2015

### HallsofIvy

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

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

7. Apr 2, 2015

### Staff: Mentor

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

Chet

8. Apr 5, 2015

### masterchiefo

Sorry guys I still dont understand how to do this, been reading online tutorials on how to solve ln equation with no luck.

9. Apr 6, 2015

### 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.

10. Apr 6, 2015

### 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 eachother, 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 eachother. so you have ln(e^x+1) = ln (e^x+1)

is this right ?

11. Apr 6, 2015

### masterchiefo

posted an attempt in my last post

12. Apr 6, 2015

### ehild

It is right.

13. Apr 6, 2015

### masterchiefo

awesome man thank you very much.

14. Apr 6, 2015

### Staff: Mentor

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

15. Apr 6, 2015

### masterchiefo

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

16. Apr 6, 2015

### Staff: Mentor

The definition of lny is the power to which you have to raise e to get y. Now let's substitute ex 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