Question

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

Answer 1

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.

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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.