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

##### 1 Answer

For

For

#x = -5/3#

#### Explanation:

**Real logarithms**

As a Real valued function of Real numbers,

So if:

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

then we must have:

#4x-1 = x-6#

Subtract

#3x=-5#

Divide both sides by

#x = -5/3#

Note that with this value of

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

Since this is negative, it is not in the domain

**How about Complex logarithms?**

The Complex exponential function

#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

Then we find:

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

is well defined, so