Question

If $2^{|\ln x-2|}=\frac{4}{2^{|\ln x|}}$, then complete set of values of x is

• A. [1, e]
• B. $[1, e^2]$
• C. $[e^2, \infty)$
• D. $(0, 1] \cup [e^2, \infty)$

Answer 1

Answer: (B)

$[1, e^2]$

Answer 2

The given equation is $2^{|\ln x-2|}=\frac{4}{2^{|\ln x|}}$.

The domain of the equation requires $x > 0$ for $\ln x$ to be defined.

We can rewrite the equation as: $2^{|\ln x-2|} \cdot 2^{|\ln x|} = 4$

Using the property $a^m \cdot a^n = a^{m+n}$, we have: $2^{|\ln x-2| + |\ln x|} = 4$

Since $4 = 2^2$, we can equate the exponents: $|\ln x-2| + |\ln x| = 2$

Let $y = \ln x$. The equation becomes: $|y-2| + |y| = 2$

This equation involves the sum of two absolute values. We can solve this by considering cases based on the values of $y$ where the expressions inside the absolute values change sign, i.e., $y=0$ and $y=2$.

Case 1: $y < 0$ In this case, $y-2 < 0$ and $y < 0$. So, $|y-2| = -(y-2) = 2-y$ and $|y| = -y$. The equation becomes $(2-y) + (-y) = 2$, which simplifies to $2 - 2y = 2$. Subtracting 2 from both sides gives $-2y = 0$, so $y = 0$. This solution $y=0$ contradicts the condition $y < 0$. Thus, there are no solutions in this case.

Case 2: $0 \le y < 2$ In this case, $y-2 < 0$ and $y \ge 0$. So, $|y-2| = -(y-2) = 2-y$ and $|y| = y$. The equation becomes $(2-y) + y = 2$, which simplifies to $2 = 2$. This is an identity, which means that the equation is true for all values of $y$ in the interval $[0, 2)$.

Case 3: $y \ge 2$ In this case, $y-2 \ge 0$ and $y \ge 0$. So, $|y-2| = y-2$ and $|y| = y$. The equation becomes $(y-2) + y = 2$, which simplifies to $2y - 2 = 2$. Adding 2 to both sides gives $2y = 4$, so $y = 2$. This solution $y=2$ satisfies the condition $y \ge 2$. Thus, $y=2$ is a solution.

Combining the solutions from all cases, the values of $y$ that satisfy $|y-2| + |y| = 2$ are all $y$ such that $0 \le y < 2$ or $y=2$. This is the interval $[0, 2]$. So, $0 \le y \le 2$.

Alternatively, we can interpret $|y-2| + |y|$ geometrically as the sum of the distances from $y$ to 2 and from $y$ to 0 on the number line. The equation $|y-2| + |y| = 2$ means that the sum of the distances from $y$ to 0 and 2 is equal to the distance between 0 and 2 (which is $|2-0|=2$). This is true if and only if $y$ lies on the line segment between 0 and 2, inclusive. Thus, $0 \le y \le 2$.

Now we substitute back $y = \ln x$: $0 \le \ln x \le 2$

To solve for $x$, we exponentiate the inequality using base $e$. Since $e > 1$, the exponential function $e^u$ is increasing, so the inequality signs are preserved: $e^0 \le e^{\ln x} \le e^2$ $1 \le x \le e^2$

We must also consider the domain of the original equation, which is $x > 0$. The interval $[1, e^2]$ is a subset of $(0, \infty)$, so all values in this interval are valid solutions.

The complete set of values of $x$ is $[1, e^2]$.