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

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Answer 1 · 2015-11-14T02:34:26

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

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# power):

#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 #1/2# to both sides:

#x=(1±sqrt(4e+1))/2#

Eliminate the negative answer (in #log_"a"b, b>0#):

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