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

1 Answer
Mar 20, 2018

x=(2e)/(1-e)x=2e1e

Explanation:

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

lnx-ln(x+2)=ln(x/(x+2))=1lnxln(x+2)=ln(xx+2)=1

Now take the exponential of both sides.

e^ln(x/(x+2))=x/(x+2)eln(xx+2)=xx+2 and e^1=ee1=e so

x/(x+2)=exx+2=e

Multiply both sides by x+2x+2

x=e(x+2)x=e(x+2)

Use the distributive property.

x=ex+2ex=ex+2e

Subtract exex from both sides and factor the left hand side.

x(1-e)=2ex(1e)=2e

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

x=(2e)/(1-e)x=2e1e.