Question

solve the equation for x. lnx+ln(x-1)=ln1 and (lnx^2)-1=lnx

Answer 1

1) http://www.wolframalpha.com
2) type "(lnx^2)-1=lnx"
3) click "show steps"
4) repeat for ln(x) + ln(x-1)

Lots of practice problems here: http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm

and great videos here: http://www.mathtv.com

Hope this helps! :)

Answer 2

ln a + ln b = ln ab
ln a^2 = 2 ln a

remember the basic logarithmic properties to solve these equations

Answer 3

For the first equation:
\[\ln (1)=0\]therefore

\[\ln (x) + \ln (x-1) = 0\]and
\[\ln (x) = -\ln(x-1)\]and if you take e to the power of each side you can cancel the natural logarithms which leaves you with
\[x=\frac{1}{x-1}\]and
\[x(x-1)=1\]
\[x^2-x-1=0\]
\[Quadradic Formula: \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]