Question

Number of real solutions of the equation $\sqrt{log_{10}(-x)} = log_{10} \sqrt{x^2}$ is:

• A. none
• B. exactly 1
• C. exactly 2
• D. 4

Answer 1

Answer: (C)

exactly 2

Answer 2

1. Domain of LHS: For $\sqrt{log_{10}(-x)}$ to be defined, we need $log_{10}(-x) \ge 0$ and $-x > 0$.

   • $-x > 0 \implies x < 0$.
   • $log_{10}(-x) \ge 0 \implies -x \ge 10^0 \implies -x \ge 1 \implies x \le -1$. The intersection of $x < 0$ and $x \le -1$ gives the domain for LHS as $x \le -1$.

2. Domain of RHS: For $log_{10} \sqrt{x^2}$ to be defined, we need $\sqrt{x^2} > 0$.

   • $\sqrt{x^2} = |x|$, so $|x| > 0 \implies x \neq 0$. The domain for RHS is $x \in \mathbb{R} \setminus \{0\}$.

3. Common Domain: The common domain for the equation is the intersection of $x \le -1$ and $x \neq 0$, which is $x \le -1$.

4. Simplification: For $x \le -1$, we have $-x > 0$. Also, $\sqrt{x^2} = |x| = -x$ because $x$ is negative. The equation $\sqrt{log_{10}(-x)} = log_{10} \sqrt{x^2}$ becomes: $\sqrt{log_{10}(-x)} = log_{10}(-x)$

5. Solving the Simplified Equation: Let $y = log_{10}(-x)$. Since $x \le -1$, $-x \ge 1$, which implies $y = log_{10}(-x) \ge log_{10}(1) = 0$. The equation in terms of $y$ is: $\sqrt{y} = y$ Squaring both sides (which is valid since $y \ge 0$): $y = y^2$ $y^2 - y = 0$ $y(y-1) = 0$ This yields two possible values for $y$: $y=0$ or $y=1$. Both are $\ge 0$, so they are valid solutions for $\sqrt{y}=y$.

6. Finding x values:

   • If $y=0$: $log_{10}(-x) = 0 \implies -x = 10^0 = 1 \implies x = -1$. This value $x=-1$ is in the common domain $x \le -1$.
   • If $y=1$: $log_{10}(-x) = 1 \implies -x = 10^1 = 10 \implies x = -10$. This value $x=-10$ is in the common domain $x \le -1$.

7. Conclusion: There are two distinct real solutions: $x=-1$ and $x=-10$.