# How do you find the domain of C(x)=ln((x+3)^4 )?

Sep 14, 2016

The domain of $C \left(x\right)$ is $\left(- \infty , - 3\right) \cup \left(- 3 , \infty\right)$

#### Explanation:

Assuming we're dealing with the Real natural logarithm, the domain is $\mathbb{R} \text{\} \left\{- 3\right\}$

When $x = - 3$:

${\left(x + 3\right)}^{4} = {\left(\left(- 3\right) + 3\right)}^{4} = {0}^{4} = 0$

and $\ln \left(0\right)$ is (always) undefined.

So $- 3$ is not in the domain.

When $x \ne - 3$

${\left(x + 3\right)}^{4} > 0$

so $\ln \left({\left(x + 3\right)}^{4}\right)$ is well defined.

So the domain is the whole of the Real numbers except $- 3$.

In interval notation $\left(- \infty , - 3\right) \cup \left(- 3 , \infty\right)$

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Footnote

$C \left(x\right) = \ln \left({\left(x + 3\right)}^{4}\right)$

into:

$C \left(x\right) = 4 \ln \left(x + 3\right)$

While this would be true for any $x > - 3$, it is not true for $x < - 3$.

In fact, if you extend the definition of $\ln$ to Complex values, allowing the logarithm of negative numbers, then if $t < 0$ we have:

$\ln t = \ln \left\mid t \right\mid + \pi i \text{ }$ (principal value)

and hence if $x < - 3$:

$4 \ln \left(x + 3\right) = 4 \left(\ln \left\mid x + 3 \right\mid + \pi i\right) = 4 \ln \left\mid x + 3 \right\mid + 4 \pi i \ne 4 \ln \left\mid x + 3 \right\mid = \ln \left({\left(x + 3\right)}^{4}\right)$