Question

solve the following

ln(x+1)=2+ln(x-1)

Answer 1

Do you know your log rules? how about ln(x)-ln(y) ?

Answer 2

yea

Answer 3

Cool, lets start by subtracting the ln(x-1) to the other side of the equation. Then use the log rule that you know from above. Write out the new equation using the equation editor and then we'll go from there. :D

Answer 4

can u draw it

Answer 5

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

Answer 6

\[\ln(x+1)-\ln(x-1)=2\]

Answer 7

Do you know how to "cancel" out a natural log?

Oh and yes you are correct. My equation then takes the log rule i asked if you knew from above and applied it. Does that make sense?

Answer 8

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

Answer 9

Nice! now you just have to solve for x. :D

Answer 10

what happens to the e

Answer 11

e is just a number. thing of it as such. So e^2 will just become apart of your answer. Post back your answer and i'll check to see if it's correct. :D

Answer 12

Err Think, not thing. sorry.

Answer 13

(x+1)=e^2(x−1)

Answer 14

Yup, that's what i'd do so far. Now distribute the e^2. My hint to solve x will be to subtract it over after you distribute and then to factor. Let me know what you get.

Answer 15

\[x+1=e ^{2}x-e ^{2} \]

Answer 16

still there

Answer 17

I am. But you have no intention of trying to learn this on your own. You've made this very clear by your last reply to another question. Therefor, i'm deciding not to engage myself in helping you any further.

Answer 18

Lets try this again. :D
So your last equation is correct. so if you subtract e^2*x over you get:
\[x+1-e^2*x=-e^2\]
subtract the 1 over
\[x-e^2*x=-e^2 - 1\]
Then factor the x
\[x(1-e^2)=-1-e^2\]

I'm sure you can solve it from there.