How to find range of a logarithmic function? 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.
 A: 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 $\ln(\frac{11}3)$.
A: 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 $\langle \leftarrow , 2]$. 
A: Hint: 
In the reals $\log_{b}(p(x))$ is not defined when $p(x) \leq 0$. Where $p(x)$ is a polynomial.
