# Domain of $\;y =\; \frac{\sqrt {4 - \log_2 x}}{\log_2 (x - 1)}$ [closed]

I am having some difficulty in finding the domain of this function:

$$y = \frac{\sqrt {4 - \log_2 x}}{\log_2 (x - 1)}$$

Any help?

• I am observing thus far, four answers to a question that were all posted within a 5 minutes "radius" starting as early as 6 minutes after the question is posted, all this, for an asker that has disregarded and failed to provide the context for this question, any thoughts on the question, let alone any work by the asker. Dec 16, 2016 at 17:59
• While I mostly just lurk these days, the behavior you (@amWhy) describe is a sure way to get a down vote from me. I won't do it to these answers, because identifying yourself as a down-voter is a sure way to be targeted for retaliation! Dec 16, 2016 at 18:12
• @TheChaz2.0 I miss seeing you around! Dec 16, 2016 at 19:38

HINTS

You need $x-1 >0$ for $\log_2(x-1)$, and hence $\log_2 x$, to be defined.

You need $\log_2(x-1) \neq 0$ for division by $\log_2(x-1)$ to be defined.

You need $4-\log_2x \ge 0$ for $\sqrt {4-\log_2x}$ to be defined.

You need all of the above for your function to be defined.

Note the following three things:

• $\log(n)$ is only defined when $n > 0$.
• $\sqrt{n}$ is only defined when $n \ge 0$.
• $\frac ab$ is only defined when $b\ne0$.

For $f(x)$ to exist, the numerator needs to exist, and the denominator needs to exist and be non-zero.

• The numerator exists if the expression under the square root exists, and is is $\geq 0$.

• $4-\log_2 x$ exists if $x > 0$; it’s nonnegative if $4 \geq \log_2 x$, or $x \leq 16$.
• The denominator exists if $x - 1 > 0$, so $x > 1$. It equals zero when $x=2$, so we require $x \neq 2$.

Our combined requirements are $x \leq 16$ and $x > 0$ and $x > 1$ and $x \neq 2$. So the domain is $(1, 16] \setminus \{2\}$.

You must have

• $x>0$

• $\ln(x)\leq4\ln(2)$ or $x\leq 16$

• $x>1$

• $x-1\neq 1$ or $x\neq 2$

thus the domain of definition is

$$D_f=(1,2)\cup(2,16].$$

• How did you arrive at your four bullet points? Dec 16, 2016 at 18:12

$\displaystyle Y(x) = \frac{\sqrt {4 - \log_2 x}}{\log_2 (x - 1)}$

For the $\log_2(a)$ you need $a>0$, so for the numerator $x>0$ and for the denominator $(x-1)>0\iff x>1$.

Also you need the denominator not to be zero. We know $\log_2(a)=0\iff a=1$ so we need $(x-1)\neq 1\iff x\neq2$

Last part for $\sqrt a$ we need $a\ge0$ so $4-\log_2(x)\ge 0\iff \log_2(x)\le 4\iff x\le 2^4\iff x\le 16$ .

Finally, if we put all this together $x\in ]1,16]\backslash\{2\}$

• Maybe you'd like to edit your mention of $log_2$ so that it looks normal, like $\log_2$.? For future reference, write \log_a and not log_a. That gets you $\ln_a$ vs $ln_a$ Dec 16, 2016 at 18:08
• Sorry, this is my first post, I'm not yet used to all LaTeX details. Thanks for having edited it.
– zwim
Dec 16, 2016 at 18:13
• No apologies needed when you're learning to fine-tune your formatting. Dec 16, 2016 at 18:15