Question

How to simplify e^[(-1)((2x+1)+lnabs(2x+1)+C)]?

Answer 1

\[e ^{-1[(2x+1)+\ln \left| 2x+1 \right|+C]}\]

Answer 2

\[K\frac{ 2x+1}{ e ^{2x+1}}\] @Idealist10

Answer 3

that's only

Answer 4

e^c = k

Answer 5

@Idealist10 did u understand how i do it

Answer 6

How did you get (2x+1)/(e^(2x+1))?

Answer 7

\[e^{\ln(2x+1)}= 2x+1\]
\[e ^{-(2x+1)}=\frac{ 1 }{ e ^{2x+1}}\]

Answer 8

so ?

Answer 9

But \[e ^{-\ln \left| 2x+1 \right|}\]
is still 2x+1? Not 1/(2x+1)?

Answer 10

no itd 1/2x+1

Answer 11

\[e^{- \ln|2x+1|}=e^{\ln|(2x+1)^{-1}|}=e^{\ln|\frac{1}{2x+1}|}=|\frac{1}{2x+1}|=\frac{1}{|2x+1|}\]