Question

ln(x+1)-ln(x-1)=3

Answer 1

Rules 2,
http://mathscene.co.uk/t/F/Su50k02_m04x.gif

Answer 2

\[ln(x+1)-ln(x-1)=3 \]\[ln\frac{(x+1)}{(x-1)}=3 \]\[\frac{(x+1)}{(x-1)}=e^3 \]\[(x+1)=e^3 (x-1)\] Can you solve it now?

Answer 3

Do I distribute? or divide by (x+1) ? I'm not very good at solving logarithmic equations with the natural log :(

Answer 4

Use a calc to solve e^3

Answer 5

e^3 = 20

Answer 6

Hmmm...... personally I wouldn't do so :|
\[x+1 = e^3(x-1)\]\[x+1 = e^3x-e^3\]\[x -e^3x= -1-e^3\]\[x(1 -e^3)= -(1+e^3)\]\[x= -\frac{(1+e^3)}{(1 -e^3)} = \frac{(1+e^3)}{(e^3-1)}\]

Answer 7

\[\large \begin{align} &(x+1) = e^3(x-1)\\ & x+1= e^3 x - e^3\\ & x - e^3 x = -1 - e^3\\ & x(1 - e^3) = -(1 + e^3)\\ & x = -\frac{1 + e^3}{1 - e^3} \end{align}\]

Answer 8

lol calli typed faster :P