Finding the domain (and range) of a logarithmic function

[brotherbobby]

Homework Statement

Find the domain (and range) of $y = \log_{(x-4)}(x^2-11x+24)$

Relevant Equations

Given the function $y = \log_a x$, we have the following domain $x>0$ and the base $a>0\quad (a\ne 1)$. The range of the function $y\in \mathbb{R}$.

Problem statement : I copy and paste the problem from the text. You will note that I added the range myself, because it seemed relevant and yet I couldn't do it.


Attempt : I could evaluate the domain.

The base of the function $x-4>0\Rightarrow x>4.\hspace{60 pt} (1)$

The function $x^2-11x+24>0\Rightarrow (x-3)(x-8)>0\Rightarrow x<3\quad \rm{or}\quad x>8\hspace{60 pt} (2).$

From $(1)\;\rm{and}\;(2)$, we have to reject $\cancel{x<3}$. Also $(1)\;\rm{and}\;(2)$ must hold simultaneously, similar to an $\rm{AND}$ in between them. That leaves us with $\boxed{x>8}$ as the domain, or writing it nicely : $\text{Domain of}\; y\mapsto \boxed{x\in(8,\infty)}.\quad \huge{\color{green}{\checkmark}}$

This agrees with the solution in the text. The typing is poor and incoherent at places. For one, I didn't understand why the authour said $x\ne 5$, though of course it isn't. That way, $x\ne 6$ too.


Doubt : $\qquad\textsf{How to find the range?}$

The given function is $y = \log_{(x-4)}(x^2-11x+24)$. We know that $x>8$. Let me put $x=8+\epsilon, \quad(\epsilon>0)$. We have $y = \log_{\epsilon+4}[\epsilon(\epsilon+5)]$. How does one evaluate this in the limit $\epsilon\to 0$?

I tried with a two values of $\epsilon=10^{-3}, 10^{-6}$. I got answers of $-3.8, -8.8$ respectively.

Evaluating online, I find that the limit does not exist, presumably because it's $-\infty$?

I have put $\epsilon+4 = x$.

On the other extreme, it seems that the limit is 2.








Here is the graph from $\texttt{Desmos}$. The range of the function turns out to be $\boxed{y\in(-\infty, 2)}$

Request : A hint as to how to evaluate the limit of the function $\displaystyle{\lim_{x\to 0} \log_x (x-4)(x+1)}$.

 

[PeroK]

You really want to calculate the limit as $x \to 8^+$.

 

[brotherbobby]

Yes.

 

[PeroK]

Yes.

The quadratic tends to 0 and the base of the log tends to 4. The limit must be $-\infty$. It depends how rigorous you need to be.

 

[brotherbobby]

Thank you. Sorry should have done that myself. $4^x\to 0\Rightarrow x\to -\infty$.

Do you have a hint as to how to calculate the upper limit as $x\to +\infty$?

 

[PeroK]

Thank you. Sorry should have done that myself. $4^x\to 0\Rightarrow x\to -\infty$.

Do you have a hint as to how to calculate the upper limit as $x\to +\infty$?

You could differentiate the function and show it has no turning points. The limit of $2$ should be easy enough. $\log ab = \log a +\log b$

 

[pasmith]

Homework Statement: Find the domain (and range) of $y = \log_{(x-4)}(x^2-11x+24)$

...

For one, I didn't understand why the authour said $x\ne 5$,

When x = 5, the base of the logarithm is 1. Is that possible?

Alternatively, you can use <br /> \log_{(x-4)} f(x) = \frac{\ln f(x)}{\ln (x - 4)} and the denominator vanishes when x = 5.

 

[PeroK]

With the function in that form, it's straightforward to show, without differentiation, that $y < 2$.

 

[brotherbobby]

You could differentiate the function and show it has no turning points. The limit of 2 should be easy enough.

I'd like to take this a step at a time. Let me try differentiating the function first. Since that is proving difficult, as you will see, finding the limit will have to wait.

The given function is $y = \log_{(x-4)}(x^2-11x+24)=\dfrac{\ln(x^2-11x+24)}{\ln(x-4)}$.

Necessary information : That $\frac{d}{dx}\ln x=\frac{1}{x}$, along with the $\dfrac{u}{v}$ and the $\rm{chain}$ rules. For a turning point at some $x=x_0\;$, $\left(\dfrac{d}{dx}\right)_{x=x_0}f(x)=0$.

Thus
\begin{gather*}
\dfrac{d}{dx}\dfrac{\ln(x^2-11x+24)}{\ln(x-4)}=\displaystyle{\dfrac{{\dfrac{(2x-11)\ln(x-4)}{(x-3)(x-8)}-\dfrac{\ln[(x-3)(x-8)]}{x}}}{[\ln(x-4)]^2}}\\
= \dfrac{x(2x-11)\ln(x-4)-(x-3))(x-8)\ln(x-3)(x-8)}{x(x-3)(x-8)[\ln(x-4)]^2}
\end{gather*}
I have to show that this expression is never zero. Remembering the domain where $\boldsymbol{x>0}$, I find that the demoninator is always greater than 0.
For the numerator, it should turn out that $x(2x-11)\ln(x-4)>(x-3)(x-8)\ln(x-3)(x-8)$ for $x>8$.

This is where am stuck. How do I show that the first expression is greater than the second?

Many thanks.

 

[PeroK]

I think the best approach is to let $z = x-4$, so we have (for $z > 4$):
$$y = \ln_z\big ( (z+1)(z-4)\big) = \frac{\ln\big ( (z+1)(z-4)\big)}{\ln z} = \frac{\ln(z+1) + \ln (z-4)}{\ln z}$$First, we can see that as $z \to 4^+$, $\ln z$ and $\ln(z+1)$ are positive and bounded and $\ln(z-4) \to -\infty$. Hence as $z \to 4^+$, $y \to -\infty$.

As $z \to \infty$, we have $\frac{\ln(z+1)}{\ln z} \to 1$ and $\frac{\ln(z-4)}{\ln z} \to 1$, so that $y \to 2$.

It only remains to show that $y < 2$ for all $z > 4$. You can do this by differentiation to show that the function has no turning points, hence is increasing. Or, as the function is continuous, by the intermediate value theorem, it is enough to show that $y \ne 2$.

If $y = 2$, then: $$\ln\big ( (z+1)(z-4)\big) = 2\ln z = \ln z^2$$Hence:$$z^2 - 3z - 4 =z^2$$This is not possible for $z > 4$, hence the range of $y$ is $(-\infty, 2)$.

 

[pasmith]

You also have the series expansion, valid for z &gt; 4, \begin{split}<br /> \frac{\ln(z+1) + \ln(z - 4)}{\ln z} &amp;= 2 + \frac{1}{\ln z}\left( \ln\left(1 + \frac 1z\right) + \ln\left(1 - \frac4z\right)\right) \\<br /> &amp;= 2- \frac{1}{\ln z} \sum_{n=1}^\infty \frac{4^n + (-1)^n}{n} \frac{1}{z^n} \end{split} where every term of the power series is positive. This shows both that the function is monotonic increasing (since z^n \ln z is monotonic increasing for z &gt; e^{-1/n}), and that its supremum is 2.

 

[pasmith]

Another alternative is to note that if \log_z(z + 1)(z - 4) = y then <br /> (z + 1)(z - 4) - z^y = 0. Setting t = 1/z for t \in (0, \frac14) and multiplying by t^2 gives <br /> g(t) = (1 + t)(1 - 4t) - t^{2-y} = 0. For y &gt; 2, t^{2 - y} &gt; 1 and 0 &lt; (1 + t)(1 - 4t) &lt; 1 so that g(t) &lt; 0 and there is no solution. For y = 0 the only solutions are t = 0 and t = -3/4, neither of which are in (0, \frac14). For y &lt; 2, g(0) = 1 &gt; 0 and g(\frac14) = -(\frac 14)^{2 - y} &lt; 0 so that by the intermediate value theorem there exists t \in (0, \frac14) such that g(t) = 0.