## anonymous 5 years ago what is the natural log of e^-x? Can anyone help?

1. anonymous

-x

2. anonymous

$ln(e^x)=x$ because the log is the inverse of the exponential.

3. anonymous

likewise $e^{ln(x)}=x$

4. anonymous

how do i solve this one then? Do you mind helping? it's e^x - 11e^-x. I need to set it to 0 and solve for x.

5. anonymous

ok let me write it.

6. anonymous

thanks.

7. anonymous

ok i think the simplest thing to do is remember that $e^{-x}=\frac{1}{e^x}$

8. anonymous

actually first lets do a simple trick. multiply by $e^x$ to get $e^x(e^x-11e^{-x})=0$ $e^{2x}-11e^{-x+x})=0$ $e^{2x}-11e^0=0$ $e^{2x}-11=0$

9. anonymous

hope that was not too many steps i just wanted to make sure it was clear. now we solve for x: $e^{2x}=11$ $2x=ln(11)$ $x=\frac{ln(11)}{2}$

10. anonymous

thank you thank you thank you in advance for taking the time!

11. anonymous

if any steps are not clear let me know. btw we can multiply by $e^x$ because it is never 0. in fact $e^x$ is always positive.

12. anonymous

thankyou! what about solving e^x + e^-x - 12 for x? I jusgt really don't get this stuff. Your help is wonderful!

13. anonymous

is that $e^x+e^{-x}=12$?

14. anonymous

it's e^x + e^-x -12 = 0

15. anonymous

try multiplying again by $e^x$ and see what happens. i will do it with pencil because if i type on the fly i might make a mistake

16. anonymous

do it the same way again then? It's so funny that this is a question on our test and we have never done anything like it! Is that calc one? That is way more than I am ready for after only 3 weeks of calculus one.

17. anonymous

well this one is a pain. first of all multiply by $e^x$

18. anonymous

get $e^{2x}+1-12e^x=0$ or $e^{2x}-12e^x+1=0$

19. anonymous

this one in a quadratic equation in $e^x$ and it doesn't factor so you have to use the quadratic formula with a = 1, b = -12 and c = 1

20. anonymous

hope it is clear that it is a quadratic.

21. anonymous

maybe i messed up. oh no. ok here goes.

22. anonymous

$e^{2x}-12e^x+1=0$ quadratic equation in $e^x$ so solve $z^2-12z+1=0$ using the quadratic formula and then replace z by $e^x$

23. anonymous

i just want to know how yhou know all this stuff?

24. anonymous

$z^2-12z+1=0$ $z^2-12z=-1$ $(z-6)^2=35$ $z-6=\pm\sqrt{35}$ $z=6\pm \sqrt{35}$

25. anonymous

oh i didn't use the quadratic formula i completed the square. same thing but it keeps me from having to simplify the radical

26. anonymous

you will get the same answer with formula, it will just take longer.

27. anonymous

omg thanks again. i am writing frantically to keep up, but i am not even close to your abilities!

28. anonymous

$e^{x}=6+\sqrt{35}$ $x=ln(6+\sqrt{35})$ \ $x=ln(6-\sqrt{35})$

29. anonymous

well it is written here. if you have any questions about any steps ask

30. anonymous

btw if you didn't know the trick of multiplying by $e^x$ you could have added: $e^x+e^{-x}=12$ $e^x+\frac{1}{e^x}=12$ $\frac{e^{2x}+1}{e^x}=12$ $e^{2x}+1=e^{12x}$ and you see you get the same equation again.

31. anonymous

Again, thanks and maybe I will talk to you again soon if you have another 3 hours for another problem! : )

32. anonymous

good luck with calc!