what is the natural log of e^-x? Can anyone help?
-x
\[ln(e^x)=x\] because the log is the inverse of the exponential.
likewise \[e^{ln(x)}=x\]
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.
ok let me write it.
thanks.
ok i think the simplest thing to do is remember that \[e^{-x}=\frac{1}{e^x}\]
actually first lets do a simple trick. multiply by \[e^x\] to get \[e^x(e^x-11e^{-x})=0\] \[e^{2x}-11e^{-x+x})=0\] \[e^{2x}-11e^0=0\] \[e^{2x}-11=0\]
hope that was not too many steps i just wanted to make sure it was clear. now we solve for x: \[e^{2x}=11\] \[2x=ln(11)\] \[x=\frac{ln(11)}{2}\]
thank you thank you thank you in advance for taking the time!
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.
thankyou! what about solving e^x + e^-x - 12 for x? I jusgt really don't get this stuff. Your help is wonderful!
is that \[e^x+e^{-x}=12\]?
it's e^x + e^-x -12 = 0
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
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.
well this one is a pain. first of all multiply by \[e^x\]
get \[ e^{2x}+1-12e^x=0\] or \[e^{2x}-12e^x+1=0\]
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
hope it is clear that it is a quadratic.
maybe i messed up. oh no. ok here goes.
\[e^{2x}-12e^x+1=0\] quadratic equation in \[e^x\] so solve \[z^2-12z+1=0\] using the quadratic formula and then replace z by \[e^x\]
i just want to know how yhou know all this stuff?
\[z^2-12z+1=0\] \[z^2-12z=-1\] \[(z-6)^2=35\] \[z-6=\pm\sqrt{35}\] \[z=6\pm \sqrt{35}\]
oh i didn't use the quadratic formula i completed the square. same thing but it keeps me from having to simplify the radical
you will get the same answer with formula, it will just take longer.
omg thanks again. i am writing frantically to keep up, but i am not even close to your abilities!
\[e^{x}=6+\sqrt{35}\] \[x=ln(6+\sqrt{35})\] \ \[x=ln(6-\sqrt{35})\]
well it is written here. if you have any questions about any steps ask
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.
Again, thanks and maybe I will talk to you again soon if you have another 3 hours for another problem! : )
good luck with calc!
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