Question

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

• A. zero
• B. exactly 1
• C. exactly 2

Answer 1

Answer: (C)

exactly 2

Answer 2

The given equation is $\log_{10} \sqrt{\log_{10} |x|-x} = \log_{10} \sqrt{x^2}$.

The domain requires:

1. $\log_{10} |x|-x > 0$
2. $x^2 > 0 \implies x \neq 0$

Equating the arguments of the logarithm: $\sqrt{\log_{10} |x|-x} = \sqrt{x^2}$ $\log_{10} |x|-x = x^2$ $\log_{10} |x| = x^2 + x$

Case 1: $x > 0$ $|x| = x$. The equation becomes $\log_{10} x = x^2 + x$. The domain condition is $\log_{10} x - x > 0$. Let $k(x) = \log_{10} x - x$. $k'(x) = \frac{1}{x \ln 10} - 1$. Setting $k'(x)=0$ gives $x = \frac{1}{\ln 10}$. The maximum value of $k(x)$ is $k(\frac{1}{\ln 10}) = \log_{10}(\frac{1}{\ln 10}) - \frac{1}{\ln 10} < 0$. Thus, $\log_{10} x - x > 0$ is never satisfied for $x>0$. No solutions in this case.

Case 2: $x < 0$ $|x| = -x$. Let $y = -x$, so $y > 0$. The equation becomes $\log_{10} y = (-y)^2 + (-y) \implies \log_{10} y = y^2 - y$. The domain condition becomes $\log_{10} y - (-y) > 0 \implies \log_{10} y + y > 0$.

We need to solve $\log_{10} y = y^2 - y$ for $y>0$ and check $\log_{10} y + y > 0$. Consider $f(y) = y^2 - y - \log_{10} y$. $f''(y) = 2 + \frac{1}{y^2 \ln 10} > 0$ for $y>0$, so $f(y)$ is convex. $f(1) = 1^2 - 1 - \log_{10} 1 = 0$. So $y=1$ is a solution. For $y=1$, $\log_{10} 1 + 1 = 1 > 0$. So $x=-1$ is a solution.

As $y \to 0^+$, $f(y) \to +\infty$. Since $f(y)$ is convex and has a root at $y=1$, and $f(y) \to +\infty$ as $y \to 0^+$, there must be another root $y_{root2}$ in $(0,1)$. For this root $y_{root2}$, we have $\log_{10} y_{root2} = y_{root2}^2 - y_{root2}$. The domain condition is $\log_{10} y_{root2} + y_{root2} = (y_{root2}^2 - y_{root2}) + y_{root2} = y_{root2}^2$. Since $y_{root2} > 0$, $y_{root2}^2 > 0$. So the domain condition is satisfied. Thus, there are two solutions for $y$: $y=1$ and $y=y_{root2}$, which correspond to two solutions for $x$: $x=-1$ and $x=-y_{root2}$.

Therefore, there are exactly 2 real solutions.