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The logarithmic function is not defined for any value less then or equal to zero.

Here, if $y=\log_e(x−4)$ is logarithmic function then its domain is $x \in (4, +\infty)$, and range is what we get output from given input. Here, its range is $y \in R$. That means all real numbers. It can be positive or negative. My question is for which values of $x$, $y$ becomes negative.

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    $\begingroup$ You should clarify the base of the logarithm. Since you didn't specify, I'll assume you mean the base $e$ logarithm., that is $e^x = y \iff x = \ln y$. Note that $\ln x$ is a monotonically increasing function, that $\ln 1 = 0$ that $\ln x > 0$ when $x>1$ and that $\ln x < 0$ when $x<1$. $\endgroup$ – JMoravitz Oct 18 '18 at 3:06
  • $\begingroup$ would you please elaborate how X<1 and above function still then defined. $\endgroup$ – Rakibul Islam Oct 18 '18 at 3:22
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    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 18 '18 at 10:00
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The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = e^x$ is a strictly increasing function with $y$-intercept $f(0) = 1$ that assumes only positive values. Thus, for each $x < 0$, $0 < f(x) < 1$, as shown below.

graph_of_exponential_function

Since $\lim_{x \to -\infty} f(x) = 0$, the line $y = 0$ (the $x$-axis) is the horizontal asymptote of the graph.

The range of $f$ is the set of all positive real numbers, that is, $\text{Ran}_f = (0, \infty)$.

If we restrict the codomain of $f$ to its range, we can define a new function $g: \mathbb{R} \to (0,\infty)$ by $g(x) = e^x$. Since $g$ is strictly increasing, for each $y > 0$, there is exactly one value of $x$ such that $g(x) = y$. By definition, $e^x = y \iff y = \log_e x$. Thus, $g$ has an inverse function $h: (0, \infty) \to \mathbb{R}$ defined by $h(x) = \log_e x$. The graph of the function $h$ is obtained from the graph of $g$ by reflecting the graph of the function $g$ in the line $y = x$, as shown below.

graphs_of_exponential_and_natural_logarithm_functions

Since $y = 0$ is a horizontal asymptote of the graph of the function $g$, the line $x = 0$ (the $y$-axis) is a vertical asymptote of the graph of the function $h$.

Notice that since $0 < g(x) < 1$ whenever $x < 0$, $h(x) < 0$ whenever $0 < x < 1$.

The graph of the function $k: (4, \infty) \to \mathbb{R}$ defined by $k(x) = \log_e (x - 4)$ is obtained by shifting the graph of the function $h$ four units to the right. Since $x = 0$ is a vertical asymptote of the graph of the function $h$, $x = 4$ is a vertical asymptote of the graph of the function $k$, as shown below.

graph_of_horizontally_shifted_natural_logarithm_function

Since $h(x) < 0$ whenever $0 < x < 1$, $k(x) < 0$ whenever $4 < x < 5$.

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$$y = \ln(x-4)$$

You can find which values of $x$ give a negative value of $y$ algebraically.

$$y < 0 \implies \ln(x-4) < 0$$ $$e^{\ln(x-4)} < e^0$$ $$x-4 < 1 \implies x < 5$$

As you mentioned at the beginning, the domain of the function is $x \in (4, +\infty)$. Thus, $4 < x < 5$ is the inequality showing the domain of input producing a negative output.

You can remember this:

Say we have a function of the type $\log_a b = c$, which can be written as $a^b = c$.

We can make the following conclusions. (Obviously $a > 1$.) $$\color{blue}{a^b < 1 \iff b<0} \implies \color{red}{\log_{a} c < 0 \iff c < 1}$$

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Consider the graph of $\ln (x-4)$ as a transformation of the graph of $\ln(x)$. The transformation is a shift to the right by 4 units. SO to find the domain, one shifts the domain 4 units to the right and the range stays the same.

Since $\ln(x) <$ 0 for x $<$ 1, then $\ln(x-4) <$ 0 for x $<$ 1+4 = 5. You can treat the case where $\ln(x-4) >$ 0 similarly.

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  • $\begingroup$ Your final sentence is a bit cryptic. $\endgroup$ – N. F. Taussig Oct 18 '18 at 10:02
  • $\begingroup$ @N.F.Taussig So sorry, I just edited the last sentence. $\endgroup$ – Joel Pereira Oct 18 '18 at 20:52

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