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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ln(e^x)=x\] because the log is the inverse of the exponential.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0likewise \[e^{ln(x)}=x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok i think the simplest thing to do is remember that \[e^{x}=\frac{1}{e^x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually first lets do a simple trick. multiply by \[e^x\] to get \[e^x(e^x11e^{x})=0\] \[e^{2x}11e^{x+x})=0\] \[e^{2x}11e^0=0\] \[e^{2x}11=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hope 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
 5 years ago
Best ResponseYou've already chosen the best response.0thank you thank you thank you in advance for taking the time!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0thankyou! what about solving e^x + e^x  12 for x? I jusgt really don't get this stuff. Your help is wonderful!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that \[e^x+e^{x}=12\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's e^x + e^x 12 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try 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
 5 years ago
Best ResponseYou've already chosen the best response.0do 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
 5 years ago
Best ResponseYou've already chosen the best response.0well this one is a pain. first of all multiply by \[e^x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0get \[ e^{2x}+112e^x=0\] or \[e^{2x}12e^x+1=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this 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
 5 years ago
Best ResponseYou've already chosen the best response.0hope it is clear that it is a quadratic.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe i messed up. oh no. ok here goes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[e^{2x}12e^x+1=0\] quadratic equation in \[e^x\] so solve \[z^212z+1=0\] using the quadratic formula and then replace z by \[e^x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just want to know how yhou know all this stuff?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[z^212z+1=0\] \[z^212z=1\] \[(z6)^2=35\] \[z6=\pm\sqrt{35}\] \[z=6\pm \sqrt{35}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i didn't use the quadratic formula i completed the square. same thing but it keeps me from having to simplify the radical

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you will get the same answer with formula, it will just take longer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0omg thanks again. i am writing frantically to keep up, but i am not even close to your abilities!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[e^{x}=6+\sqrt{35}\] \[x=ln(6+\sqrt{35})\] \ \[x=ln(6\sqrt{35})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well it is written here. if you have any questions about any steps ask

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0btw 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
 5 years ago
Best ResponseYou've already chosen the best response.0Again, thanks and maybe I will talk to you again soon if you have another 3 hours for another problem! : )
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