independanteng

Answered

2022-10-13

How do I find the range of these logarithmic functions?

$\begin{array}{rl}& \mathrm{ln}(3{x}^{2}-4x+5),\\ & {\mathrm{log}}_{3}(5+4x-{x}^{2}).\end{array}$

Answer & Explanation

SoroAlcommai9

Expert

2022-10-14Added 13 answers

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

So you need $3{x}^{2}-4x+5>0$ in the first case. Completing the square give you ${(x-\frac{2}{3})}^{2}+\frac{11}{9}$. We see that the quadratic is always greater than $\frac{11}{9}$ and goes to infinity. Therefore the range is $[\mathrm{ln}\left(\frac{11}{9}\right),\to \u27e9$

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 ${\mathrm{log}}_{3}(9)=2$, the range is $\u27e8\leftarrow ,2]$

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