# How do you solve #lnx+ln(x-1)=1#?

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Mar 26, 2017

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Using the rules of logarithms,

#ln(x)+ln(x-1)=ln(x*(x-1))=ln(x^2-x)# .

Therefore,

#ln(x^2-x)=1# .

Then, we exponentiate both sides (put both sides to the

#e^(ln(x^2-x))=e^1# .

Simplify, remembering that exponents undo logarithms:

#x^2-x=e# .

Now, we complete the square:

#x^2-x+1/4=e+1/4#

Simplify:

#(x-1/2)^2 = e+1/4 = (4e+1)/4#

Take the square root of both sides:

#x-1/2=(pmsqrt(4e+1))/2#

Add

#x=(1±sqrt(4e+1))/2#

Eliminate the negative answer (in

#=> color(blue)(x=(1+sqrt(4e+1))/2)#

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