How do you solve # ln x + ln(x-2) = 1#?

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Answer 1 · 2016-01-25T14:02:03

#x=1+sqrt(1+e)#

We can simplify the left hand side by using the logarithmic rule: #ln(a)+ln(b)=ln(ab)#

#ln[x(x-2)]=1#

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

To undo the natural logarithm, exponentiate both sides with base #e#.

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

#x^2-2x=e#

Move the #e# to the left side and solve with the quadratic equation.

#x^2-2x-e=0#

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

Factor a #4# from inside the square root, which can be pulled out as a #2#.

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

#x=1+sqrt(1+e)#

Notice that the negative root has been taken away, since for #ln(a),a>0# (if we are forgoing complex roots).