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