Question

solve for the variable ln(x+2)-ln(x-1)=e

Answer 1

sorry the question is
=1 not=e
ln(x+2)-ln(x+1)=1

Answer 2

I used the
a=log(base B)(b^a)
to do
ln(x+2)/(x-1)=ln(e)

Answer 3

but I don't know how to do
x+2/x-1

Answer 4

Am I doing the problem correctly as of right now?

Answer 5

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

From here, I use the property that \(log_{b}m = x \implies\ b^{x} = m\)
The base of ln is e, so I can turn this into:
\[\ln(\frac{ x+2 }{ x-1 }) = 1 \implies e = \frac{ x+2 }{ x-1 }\]

Answer 6

Ok... so then do we multiply both sides by x-1?
ex-e=x+2

Answer 7

Right. And then proceed to solve for x.

Answer 8

... sorry I don't have much experience with e so is it just going to be
ex-x=e+2

Answer 9

e is just a number, it just looks weird at first. But thats right. Now you can just factor x out of the left two terms.

Answer 10

wait..... won't it work as well if I did
ln(x+2/x-1)=ln(e) using the a=log((log base)^a)?

Answer 11

wait..... nvm sorry I confused myself

Answer 12

x(e-1)=e+2?

Answer 13

Yep. And then just divide by e-1 to get \(x = \frac{e+2}{e-1}\)

Answer 14

Ok I got it now. Thanks!

Answer 15

You're welcome :)

Answer 16

but how would it work for
ln(x+2)-ln(x-1)=1
x+2-(x-1)=e
(e+2/e-1)+2-(e+2/e-1)+1=e
(e+2/e-1)-(e+2/e-1)=e-3

Answer 17

or do we divide the ln() on the left side to get
(e+2/e-1)/(e+2/e-1)=e+3

Answer 18

*e-3 not e+3