Solve this separable differential equation

How would I go about solving the following separable differential equation?

$\frac{dx(t)}{dt}=8-3x$ with $x(0)=4$?

My solution thus far is the following:

$\int \frac{dx(t)}{8-3x}=\int dt+C$

$\Rightarrow \text{ln}|8-3x(t)|=-3(t+C)$

Now using the fact that $x(0)=4$ we can solve for $C=-\frac{1}{3}\text{ln}|4|$. From this, I would seem to get

$|8-3x(t)|={e}^{-3t-\frac{1}{3}\text{ln}4}$

but something seems to go wrong. Am I making a mistake somewhere and how could I solve this entirely?

How would I go about solving the following separable differential equation?

$\frac{dx(t)}{dt}=8-3x$ with $x(0)=4$?

My solution thus far is the following:

$\int \frac{dx(t)}{8-3x}=\int dt+C$

$\Rightarrow \text{ln}|8-3x(t)|=-3(t+C)$

Now using the fact that $x(0)=4$ we can solve for $C=-\frac{1}{3}\text{ln}|4|$. From this, I would seem to get

$|8-3x(t)|={e}^{-3t-\frac{1}{3}\text{ln}4}$

but something seems to go wrong. Am I making a mistake somewhere and how could I solve this entirely?