Question

what is the natural log of e^-x? Can anyone help?

Answer 1

-x

Answer 2

\[ln(e^x)=x\] because the log is the inverse of the exponential.

Answer 3

likewise \[e^{ln(x)}=x\]

Answer 4

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.

Answer 5

ok let me write it.

Answer 6

thanks.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

is that \[e^x+e^{-x}=12\]?

Answer 14

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

Answer 15

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

Answer 16

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.

Answer 17

well this one is a pain. first of all multiply by \[e^x\]

Answer 18

get
\[ e^{2x}+1-12e^x=0\] or
\[e^{2x}-12e^x+1=0\]

Answer 19

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

Answer 20

hope it is clear that it is a quadratic.

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

good luck with calc!