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

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Answer 1 · 2018-03-20T03:02:53

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

First use the property of logarithms that #ln(a)-ln(b)=ln(a/b)#.

#lnx-ln(x+2)=ln(x/(x+2))=1#

Now take the exponential of both sides.

#e^ln(x/(x+2))=x/(x+2)# and #e^1=e# so

#x/(x+2)=e#

Multiply both sides by #x+2#

#x=e(x+2)#

Use the distributive property.

#x=ex+2e#

Subtract #ex# from both sides and factor the left hand side.

#x(1-e)=2e#

Divide both sides by #(1-e)#.

#x=(2e)/(1-e)#.