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

- anonymous

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

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- anonymous

-x

- anonymous

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

- anonymous

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

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## More answers

- 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.

- anonymous

ok let me write it.

- anonymous

thanks.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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.

- anonymous

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

- anonymous

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

- anonymous

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

- 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

- 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.

- anonymous

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

- anonymous

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

- 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

- anonymous

hope it is clear that it is a quadratic.

- anonymous

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

- 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 \[e^x\]

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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.

- anonymous

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

- anonymous

good luck with calc!

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