Question

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

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

Answer 1

Answer: (A), (B)

zero

Answer 2

For the Left-Hand Side (LHS), $\log_{10} \sqrt{\log_{10}(-|x|)}$, to be defined in real numbers, the argument of the outer logarithm must be positive: $\sqrt{\log_{10}(-|x|)} > 0$. This implies $\log_{10}(-|x|) > 0$. For $\log_{10}(-|x|)$ to be defined, its argument must be positive: $-|x| > 0$. The condition $-|x| > 0$ implies $|x| < 0$. However, for any real number $x$, $|x| \ge 0$. Thus, $|x| < 0$ has no real solutions. Since the condition $-|x| > 0$ is never satisfied for any real $x$, the term $\log_{10}(-|x|)$ is never defined in real numbers. Consequently, the entire LHS expression is undefined for all real $x$.

For the Right-Hand Side (RHS), $\log_{10} \sqrt{x^2}$, to be defined in real numbers, the argument of the logarithm must be positive: $\sqrt{x^2} > 0$. Since $\sqrt{x^2} = |x|$, this condition becomes $|x| > 0$, which means $x \neq 0$. The RHS is defined for all real numbers except $x=0$.

For the original equation to have a real solution, both the LHS and the RHS must be defined for the same real value of $x$. As the LHS is never defined for any real $x$, there are no real values of $x$ for which the equation holds true. Therefore, the number of real solutions is zero.