How do you solve #Ln(4x-1) = Ln(x-6) # and find any extraneous solutions?

1 Answer
Jun 5, 2016

Answer:

For #ln# as a Real valued function of Real number this has no solutions.

For #ln# as a Complex valued function of Complex numbers:

#x = -5/3#

Explanation:

Real logarithms

As a Real valued function of Real numbers, #ln# is one-one on its domain #(0, oo)#.

So if:

#ln(4x-1) = ln(x-6)#

then we must have:

#4x-1 = x-6#

Subtract #x-1# from both sides to get:

#3x=-5#

Divide both sides by #3# to get:

#x = -5/3#

Note that with this value of #x#:

#4x-1=x-6 = -23/3#

Since this is negative, it is not in the domain #(0, oo)# and its Real logarithm is undefined.

#color(white)()#
How about Complex logarithms?

The Complex exponential function #e^z# is not one-one, but we can extend the definition of the Real logarithm to Complex values with the definition:

#ln(z) = ln(abs(z)) + i Arg(z)#

This has the property that:

#e^(ln(z)) = z# for all #z in CC#

but note that:

#ln(e^z) = z# for all #z in CC# with #Im(z) in (-pi, pi]#

For #z# with values of #Im(z)# outside this range, #ln(e^z) != z#

Then we find:

#ln(-23/3) = ln(23/3) + pii#

is well defined, so #x = -5/3# is a solution.