Question

Simplify -e^(-x)(x+1)+e^(-x)lnabs(e^x+1)+lnabs(1+e^(-x)).

Answer 1

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

Answer 2

and then apply the identity ln a + ln b = ln ab

Answer 3

i dont think hell be able to factor out an e^-x

Answer 4

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

Answer 5

no dont factor out

Answer 6

apply the identity ln a + ln b = ln ab

Answer 7

So \[(e^x+1)(1+e ^{-x})\]?
Since you said ab?

Answer 8

Btw you can take out the absolute sign since you cant input a negative number in a ln function

Answer 9

and yes

Answer 10

\[\large\rm -e^{-x}(x+1)+e^{-x}ln(e^x+1)+ln(1+e^{-x}) \]

Answer 11

\[\large\rm -e^{-x}(x+1)+e^{-x}ln((e^x+1)(1+e^{-x})) \]

Answer 12

\[ \large\rm -e^{-x}(x+1)+e^{-x}ln(e^x+1+1+e^{-x} )\]

Answer 13

Now take e^-x common simplify the bracket in ln and you will have your answer

Answer 14

@mathmale Right?

Answer 15

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

Answer 16

you can also do this (e^x+2+e^-x) = (e^0.5x + e^-0.5x)^2

Answer 17

The result I got is\[e ^{-x}[\ln(e^x+2+e ^{-x})-x-1]\]

Answer 18

\[\large\rm e^x+2+e^{-x} = (e^{0.5x} + e^{-0.5x})^2\]

Answer 19

Thank you!