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

Question

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?

Answer 1

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.

Answer 2

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 $b\ne0$.

Answer 3

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 $(1, 16] \setminus \{2\}$.

Answer 4

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

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

Answer 5

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

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 $x\in ]1,16]\backslash\{2\}$