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 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 $\ln(\frac{11}3)$.
Hint: 

In the reals $\log\_{b}(p(x))$ is not defined when $p(x) \leq 0$. Where $p(x)$ is a polynomial.