<s>[INST] 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.
 [/INST] 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 $\langle \leftarrow , 2]$. 
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