# For which values of $x$ does $y = \log_e (x - 4)$ become negative?

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.

• 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 $x<1$. – JMoravitz Oct 18 '18 at 3:06
• would you please elaborate how X<1 and above function still then defined. – Rakibul Islam Oct 18 '18 at 3:22
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Oct 18 '18 at 10:00

## 3 Answers

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.

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

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

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 $$a > 1$$.) $$\color{blue}{a^b < 1 \iff b<0} \implies \color{red}{\log_{a} c < 0 \iff c < 1}$$

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.

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