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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 18 '18 at 10:00

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.