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

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! :)

ln a + ln b = ln ab ln a^2 = 2 ln a remember the basic logarithmic properties to solve these equations

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 \[x(x-1)=1\] \[x^2-x-1=0\] \[Quadradic Formula: \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

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