Question

Solve for x
e^xe^(x+1)=1

Answer 1

\[e^{x}e^{(x+1)}=1\]

Answer 2

you can start by dividing e^x on both sides

Answer 3

then compare the exponents on both sides :)

Answer 4

wouldn't it be better to just write it as
e^(2x+1) = 1
then take the ln of both sides, the exploit the rule ln(a^b) = bln(a)

Answer 5

\(\Large \dfrac{1}{a^b} = a^{-b}\)

Answer 6

Would that cancel e^x on the left ad put it right next to the 1?

\[e^{(x+1)}=1e^{x}\]

Answer 7

thats a good approach too Austral :)

Answer 8

\(\large e^{x+1}= e^{-x}\)

Answer 9

x+1 =-x

Answer 10

\(\large\color{black}{ e^{x}e^{x+1}=1 }\)
\(\large\color{black}{ e^{x+x+1}=1 }\)
\(\large\color{black}{ e^{2x+1}=1 }\)
\(\large\color{black}{ e^{2x+1}=e^{\ln(1)} }\)
\(\large\color{black}{ 2x+1=\ln(1) }\)
and we all know what \(\large\color{black}{ \ln(1) }\) is.

Answer 11

because
\(\Large \dfrac{1}{e^x} = e^{-x}\)

Answer 12

my method got away with the log part .-.

Answer 13

there is actually no "lol" part, or not that I would say that my solution involves logarithms.

Answer 14

Opps, log part, not lol part.

Answer 15

lol

Answer 16

ria,
confused?
got it ?
start over?

Answer 17

Uhm... ln(1)=0
So... Wait I have a couple questions... :D

Yhu got 2x+1 because yhu combined the like terms right?

Answer 18

\(
\Large a^ma^n = a^{m+n}\)

so yeah
x+ (x+1)

combining like terms gives
2x+1

Answer 19

yes, added exponents of e, to get 2x+1

Answer 20

And ln=0, from there, would 2x+1 be my answer, or would I need to solve it from there like I did with the last one.
x=-2?

Answer 21

yhu need to solve

2x+1= 0

and x is NOT -2

Answer 22

.-.
-1/2?

Answer 23

yheshhhh

Answer 24

i mean thats correct :) :P

Answer 25

^o^ Yea!

Answer 26

Thank yhu both ^u^
I'm gonna medal Solomon cuz he didn't get a medal.
Thank yhu again so much Hartnn ^.^ Yhur really helpful. n~n

Answer 27

There should be one for everyone ;)
most welcome ^_^

Answer 28

I did pretty much the same as taking the natural log of both sides, but I used a rule
\(\Large\color{black}{ \color{red}{a}=e^{\ln(\color{red}{a})} }\)
you can prove this rule, if you lo=ike by taking the natural log of both sides.
\(\Large\color{black}{ \ln\color{red}{a}=\ln \left(\begin{matrix} e^{\ln(\color{red}{a})} \\ \end{matrix}\right) }\)
taking the exponent out,
\(\Large\color{black}{\ln( \color{red}{a})={\ln(\color{red}{a})}\ln(e) }\)

we know that, ( \(\Large\color{black}{\ln(e)=\log_ee=1 }\) )

\(\Large\color{black}{\ln( \color{red}{a})={\ln(\color{red}{a})} }\)

\(\Large\color{black}{ \color{red}{a}={\color{red}{a}} }\)

Answer 29

this rule helped me a lot in all of my courses:)

Answer 30

:O Ahh... Dude... Thank yhu. n_n

Answer 31

np