Simplifying $ \log_{e^2}(e^{4a}+ae^{4a}) $ Problem

Simplify logaritm:

$$ \log_{e^2}(e^{4a}+ae^{4a}) $$

preferably in a way that end result contains only natural logarithm.

 Attempt to solve 

I know few computational rules about logarithms:

$$ \log_a(xy) = \log_a(x)+\log_a(y) $$

$$ \log_a(\frac{x}{y})=\log_a{x}-\log_a(y) $$

$$ \log_a(x^n)=n\log_a(x) $$

And formula for change of basis :

$$ \log_a(x)=\frac{\log_b(x)}{\log_b(a)} $$

There is sum inside this logarithm and it appears we don't have formula for this. On wikipedia i found two formulas about summation / substraction inside logarithm:

$$ \log_b(a+c)=\log_b a+ \log_b (1+\frac{c}{a}) $$

$$ \log_b(a-c)=\log_b a+ \log_b (1-\frac{c}{a}) $$

I have no former experience / knowledge of these formulas but for now the assumption is these are correct. 

I would try to first try  change of basis to $\log_{e^2}()\rightarrow \log_e() = \ln()$

$$ \log_{e^2}(e^{4a}+ae^{4a})=\frac{\log_e(e^{4a}+ae^{4a})}{\log_e(e
^2)} $$

$$ \log_{e}(e^{4a}+ae^{4a})=\log_{e^2}(e^{4a}+ae^{4a})\cdot \log_e({e^2}) $$

$$ \log_{e}(e^{4a}+ae^{4a})=2 \cdot \log_{e^2}(e^{4a}+ae^{4a})$$

We get the original logarithm with base $e$

$$ \log_{e^2}(e^{4a}+ae^{4a})=\frac{1}{2}\ln(e^{4a}+ae^{4a}) $$

Now the sum:

$$ \frac{1}{2}\ln((a+1)e^{4a})
 $$

$$4a \frac{1}{2} \ln ((a+1)e) $$
$$ 4a \cdot \frac{1}{2} \ln (a+1) + \ln(e) $$

$$ \frac{4a}{2} \ln {(a+1)} + 1 $$

$$ 2a \ln (a+1) + 1 $$

If there is an error let me know.
You were doing fine down to $$\frac 12\ln((a+1)e^{4a})$$ but then you cannot pull out the $4a$ because it is not an exponent on $a+1$.  Instead you should do
$$\frac 12\ln((a+1)e^{4a})=\frac 12\left(\ln (a+1)+\ln(e^{4a})\right)\\=\frac 12\ln(a+1)+2a$$
$$\log_{e^2}(e^{4a}+ae^{4a})$$$$=\log_{e^2}e^{4a}(1+a)$$$$=\log_{e^2}e^{4a}+log_{e^2}(1+a)$$$$=\log_{e^2}e^{2.2a}+log_{e^2}(1+a)$$$$=\log_{e^2}(e^{2})^{2a}+log_{e^2}(1+a)$$$$={2a}+log_{e^2}(1+a)$$