How do you solve # lnx+ln(x-2)=3#?

2 Answers
Konstantinos Michailidis
Sep 10, 2015

We use the fact that #lna+lnb=lnab#

Explanation:

Hence we have that

#lnx+ln(x-2)=3=>ln(x(x-2))=3=> x(x-2)=e^3=> x^2-2x-e^3=0#

The last is a trinomial with respect to x that has the following solutions

#x_1=1-sqrt(1+e^3),x_2=1+sqrt(1+e^3)#

Now we must not forget that #x>2# so the only acceptable solution here is #1+sqrt(1+e^3)#

Gió
Sep 10, 2015

Try this:

Explanation:

We can use the rule of the logs that says:
#logx+logy=log(xy)# to get:

#ln[x(x-2)]=3#
apply the definition of log (natural, with base #e#) to get:
#x(x-2)=e^3#
rearranging:
#x^2-2x-e^3=0#

Use the Quadratic Formula to solve for #x# as:
#x_(1,2)=(2+-sqrt(4+4e^3))/2#
#x_(1,2)=(2+-sqrt(4(1+e^3)))/2#
#x_(1,2)=(2+-2sqrt(1+e^3))/2#
#x_(1,2)=1+-sqrt(1+e^3)#
Now check whether the two solutions are allowed or not by substituting them into the original equation (you'll find that only one works).

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