Question

e^5x−1⋅e^−x=2e

Answer 1

\(\large\color{black}{ e^{5x}-e^{-x}=2e }\) am I correct?

Answer 2

if you aren't going to reply, how am I supposed to help you?

Answer 3

why does that make any difference?

Answer 4

yes, I know, but what I wrote is absolutely the same thing, sint' it?

Answer 5

I guess

Answer 6

do you not know how to do this problem?

Answer 7

I am thinking that there must be a simple way to do it. I tried.

\(\large\color{black}{ e^{5x}-e^{-x}=2e }\)
\(\large\color{black}{ e^{-x}(e^{6x}-1)=2e }\)
\(\large\color{black}{ \ln[ e^{-x}(e^{6x}-1)]=\ln[2e] }\)
\(\large\color{black}{ \ln[ e^{-x}]+\ln[(e^{6x}-1)]=\ln[2e] }\)
\(\large\color{black}{ -x+\ln[(e^{6x}-1)]=\ln[2]+1 }\)

but got stuck there.

Answer 8

its a decimal

Answer 9

I want to find the exact solution first, to then approximate it.

Answer 10

I don't know how to do it. SOrry, I am confused.