- #1

brotherbobby

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- 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}##.

**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.
Problem statement : **

**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}}##

**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.

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

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)}##.

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