Is $\frac{1}{6}\ln(x+4)^6$ equivalent to $\ln(x+4)$? I was doing a question and was wondering if bringing the coefficient back into the exponent was allowed?
So, for example: $$\frac{1}{6}\ln(x+4)^6$$
Is it possible for me to do:
$$\ln(x+4)$$
Are those two equivalent?
I am struggling on what to do because I am doing a question where it is asking me to simplify to a single logarithm:
$$\frac{1}{6}\ln(x+4)^6 + \frac{1}{4}[\ln x-\ln(x^2+6x+8)^4]$$
If I can put them back into the exponent area, then I could do something such as:
$$\ln(x+4)+\ln x^\frac{1}{4}-\ln(x^2+6x+8)$$
After that, I could apply the basic adding and subtracting rules to get: $$\ln\left(\frac{(x+4)x^\frac{1}{4}}{x^2+6x+8}\right)$$
Finally, I could factor the bottom and reduce the numerator and denominator: $$\ln\left(\frac{(x+4)x^\frac{1}{4}}{(x+4)(x+2)}\right)$$
$$\ln\left(\frac{x^\frac{1}{4}}{x+2}\right)$$
Is this way of thinking right or am I not allowed to put the exponent back into the exponent area?

Let's look at some examples.  For the purposes of this discussion, all logarithms are natural (i.e., base $e$) and all logarithms are defined as functions from real numbers to the real numbers.
Consider $$f(x) = \frac{1}{6} \log (x+4)^6.$$  If $x = 0$, then we have $f(0) = \frac{1}{6} \log 4^6 = \log 4$.  But we can also choose values for which $x+4 < 0$; e.g., $$f(-5) = \frac{1}{6} \log (-5+4)^6 = \frac{1}{6} \log(-1)^6 = \frac{1}{6} \log 1 = 0.$$  In fact, the only real number for which $f(x)$ is not defined is $x = -4$.  This happens because raising $x+4$ to the sixth power always gives a nonnegative number, and in particular, a strictly positive number unless $x = -4$.
However, consider $$g(x) = \log (x+4).$$ Now if $x > -4$, we are okay because $x+4 > 0$.  But we run into problems if $x < -4$, since without the benefit of raising $x+4$ to the sixth power, it can be negative; e.g., $$g(-5) = \log(-5 + 4) = \log(-1).$$
So clearly, $f(x) \ne g(x)$ in all cases.  We do have equality if $x > -4$, but when $x < -4$, $f$ remains well-defined but $g$ does not.  To remedy this problem, we can write instead
$$h(x) = \log |x+4|,$$
and here, $f(x) = h(x)$ for all real numbers $x$ except $x = -4$, in which case neither $f$ nor $h$ are well-defined.
This suggests a path forward for your problem:  we have
$$\frac{1}{6} \log(x+4)^6  + \frac{1}{4} \left( \log x - \log (x^2 + 6x + 8)^4 \right) = \log|x+4| + \log x^{1/4} - \log |x^2 + 6x + 8|.$$  Note that we do not have to write $\log |x|^{1/4}$ (and in fact this would be incorrect), because the original function, $\frac{1}{4} \log x$, is defined if and only if $x > 0$ in the first place.  But this also means that $x+4 > 0$ and $x^2 + 6x + 8 > 0$, so it is actually the restriction imposed by $\log x$ that allows us to remove the absolute values on the other terms without loss of generality.  This gives us
$$\begin{align}\log (x+4) - \log(x^2 + 6x + 8) + \log x^{1/4} &= \log(x+4) - \log(x+4) - \log(x+2) + \log x^{1/4} \\&= \log x^{1/4} - \log (x+2) \\&= \log \frac{\sqrt[4]{x}}{x+2},\end{align}$$ where we require $x > 0$.
Keeping track of when expressions remain well-defined is an important part of simplifying algebraic expressions.
Plot of $\frac{1}{6} \log(x+4)^6 + \frac{1}{4}\left(\log x - log(x^2 + 6x + 8)^4\right)$:

Plot of $\log \frac{\sqrt[4]{x}}{x+2}$:

