Question: How do I find the range of these logarithmic functions?
\begin{align}
& \ln(3x^2 -4x +5), \\
& \log\_3(5+4x-x^2).
\end{align}
how should I approach questions like this ?

What I did: I found out the roots of quadratic: i.e. $(5-x)(1+x) >0$
so $(x-5)(1+x) <0$ so it lies between $-1$ and $5$, and then I took log on both sides of the inequality.

Answer: 
You can only take a logarithm of a number greater than zero.

So you need $3x^2-4x+5 > 0$ in the first case. Completing the square give you $\left(x-\frac{2}{3}\right)^2+\frac{11}9$. We see that the quadratic is always greater than $\frac{11}{9}$ and goes to infinity. Therefore the range is $[\ln\left(\frac{11}{9}\right), \rightarrow \rangle$

For the second one, you want $-x^2+4x+5 > 0$. We first solve $-x^2+4x+5 =0$. This gives $x=-1$ or $x=5$ as you found. Because the coefficient for $x^2$ is negative, this means that the quadratic is positive when $-1<x<5$. The maximum is attained at $x=-\frac{b}{2a}=2$, whith a value of $9$. 

So we can make the argument of the log very close to zero but never greater than 9. As $\log\_3(9)=2$, the range is $\langle \leftarrow , 2]$.
You can view the fact that as $x\to 5$ or $x\to -1$ the functional value is becoming larger and larger i.e. for $\ln(3x^2−4x+5)$ range set is unbounded. 

So, also the function is continuous on $(-1,5)$ so there must be a point in the interval where the functional value is the smallest.

Let $f(x)=\ln(3x^2−4x+5)$

$$f'(x)= \frac{6x-4}{3x^2−4x+5}$$

so at $x= \frac23$ the functional value is the lowest and is $\ln(\frac{11}3)$.