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

2 Answers
Gió
Jul 20, 2016

I found: #x=1+sqrt(1+e)#

Explanation:

We can use the property of logs that allows us to change the sum into a product of arguments as:
#ln[x*(x-2)]=1#
then use the definition of log, applied to our natural log, to write:
#x(x-2)=e^1#
#x^2-2x-e=0#
solve for #x# to get:
#x_(1,2)=(2+-sqrt(4+4e))/2=(2+-2sqrt(1+e))/2#
#x_1=1+sqrt(1+e)#
#x_2=1-sqrt(1+e)# NOT (because negative)

Costas C.
Jul 20, 2016

Use the properties:

#lna+lnb=ln(a*b)#

#lnx=y=>x=e^y#

Then solve the quadratic equation. Answer is:

#x_1=2.928#

#x_2=-0.928#

Explanation:

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

Since #lna+lnb=ln(a*b)#

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

To add an #ln# function to the right side we use #x=lne^x#

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

#lnx# is a #1-1# function:

#x*(x-2)=e#

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

Solving the quadratic:

#Δ=(-2)^2-4*1*(-e)=4+4e#

#x_(1,2)=(-(-2)+-sqrt(4+4e))/(2*1)#

From the calculator #sqrt(4+4e)=3.857#

#x_1=2.928#

#x_2=-0.928#

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