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

Jun 5, 2016

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

For $\ln$ as a Complex valued function of Complex numbers:

$x = - \frac{5}{3}$

#### Explanation:

Real logarithms

As a Real valued function of Real numbers, $\ln$ is one-one on its domain $\left(0 , \infty\right)$.

So if:

$\ln \left(4 x - 1\right) = \ln \left(x - 6\right)$

then we must have:

$4 x - 1 = x - 6$

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

$3 x = - 5$

Divide both sides by $3$ to get:

$x = - \frac{5}{3}$

Note that with this value of $x$:

$4 x - 1 = x - 6 = - \frac{23}{3}$

Since this is negative, it is not in the domain $\left(0 , \infty\right)$ 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:

$\ln \left(z\right) = \ln \left(\left\mid z \right\mid\right) + i A r g \left(z\right)$

This has the property that:

${e}^{\ln \left(z\right)} = z$ for all $z \in \mathbb{C}$

but note that:

$\ln \left({e}^{z}\right) = z$ for all $z \in \mathbb{C}$ with $I m \left(z\right) \in \left(- \pi , \pi\right]$

For $z$ with values of $I m \left(z\right)$ outside this range, $\ln \left({e}^{z}\right) \ne z$

Then we find:

$\ln \left(- \frac{23}{3}\right) = \ln \left(\frac{23}{3}\right) + \pi i$

is well defined, so $x = - \frac{5}{3}$ is a solution.