Question

ln(1+e^-x) =3 solve for x ???? cant figure these out

Answer 1

Take exp first of both sides, then

1 + e^-x = e^3

Now can you solve that for x?

Answer 2

start by writing in equivalent logarithmic form to get
\[1+e^{-x}=e^3\] and then oh again. man is the site slow

Answer 3

is it e^-x = e^3 - 1 then -x = 3-1 then -x = 2 then x = -2 sorry if thats hard to follow

Answer 4

No definitely not, because how did the -1 get into the index?

Answer 5

\[1 + e^{-x} = e^3 \implies e^{-x} = e^3 - 1\]

Answer 6

Now given that, how do you find x?

Answer 7

If I told you in general that
\[e^x = a\]
what is x?

Answer 8

is it one i am honestly not sure

Answer 9

ln x is the inverse function of e^x
Hence if e^x = a, then x = ln(a)

Answer 10

That is why when we started the problem we said that
\[\ln(1+e^-x) =3 \implies 1 + e^{-x} = e^3\]

Answer 11

make that:
\[\ln(1+e^{-x}) =3 \implies 1 + e^{-x} = e^3\]

Answer 12

because in general
\[\ln(a) = b \iff a = e^b\]

Answer 13

So now we have
\[e^{-x} = e^3 - 1\]
What do you get if you take the ln of both sides?

Answer 14

What is
\[\ln(e^{-x})\]?

Answer 15

and you're not even here. Alright. Good luck.

Answer 16

i thought that the e's were just eliminated but i am waiting on my gf to bring my textbook. looks like i have some serious studying to do

Answer 17

For the record, what is
\[\ln(e^2)\] ?