Question

Solve: $|\log_{\sqrt{3}} x - 2| + |\log_3 x - 2| = 2$

Answer 1

x = 3^{2/3}, 9

Answer 2

The given equation is $|\log_{\sqrt{3}} x - 2| + |\log_3 x - 2| = 2$. For the logarithms to be defined, we must have $x > 0$.

First, we express the logarithms in the same base. Let's use base 3. We know that $\sqrt{3} = 3^{1/2}$. Using the change of base formula $\log_{a^k} b = \frac{1}{k} \log_a b$, we have: $\log_{\sqrt{3}} x = \log_{3^{1/2}} x = \frac{1}{1/2} \log_3 x = 2 \log_3 x$.

Substitute this into the original equation: $|2 \log_3 x - 2| + |\log_3 x - 2| = 2$ $|2(\log_3 x - 1)| + |\log_3 x - 2| = 2$ $2|\log_3 x - 1| + |\log_3 x - 2| = 2$

Let $y = \log_3 x$. The equation becomes: $2|y - 1| + |y - 2| = 2$

This is an absolute value equation. We need to consider cases based on the values of $y$ where the expressions inside the absolute values change sign, i.e., the critical points $y=1$ and $y=2$.

Case 1: $y < 1$ (i.e., $y \in (-\infty, 1)$) In this case, $y - 1 < 0$ and $y - 2 < 0$. So, $|y - 1| = -(y - 1) = 1 - y$ and $|y - 2| = -(y - 2) = 2 - y$. The equation becomes: $2(1 - y) + (2 - y) = 2$ $2 - 2y + 2 - y = 2$ $4 - 3y = 2$ $3y = 2$ $y = \frac{2}{3}$ This value $y = \frac{2}{3}$ satisfies the condition $y < 1$. So, $y = \frac{2}{3}$ is a solution for $y$.

Case 2: $1 \le y < 2$ (i.e., $y \in [1, 2)$) In this case, $y - 1 \ge 0$ and $y - 2 < 0$. So, $|y - 1| = y - 1$ and $|y - 2| = -(y - 2) = 2 - y$. The equation becomes: $2(y - 1) + (2 - y) = 2$ $2y - 2 + 2 - y = 2$ $y = 2$ This value $y = 2$ does not satisfy the condition $1 \le y < 2$. So, there are no solutions for $y$ in this interval.

Case 3: $y \ge 2$ (i.e., $y \in [2, \infty)$) In this case, $y - 1 \ge 0$ and $y - 2 \ge 0$. So, $|y - 1| = y - 1$ and $|y - 2| = y - 2$. The equation becomes: $2(y - 1) + (y - 2) = 2$ $2y - 2 + y - 2 = 2$ $3y - 4 = 2$ $3y = 6$ $y = 2$ This value $y = 2$ satisfies the condition $y \ge 2$. So, $y = 2$ is a solution for $y$.

Combining the results from all cases, the possible values for $y$ are $\frac{2}{3}$ and $2$.

Now we substitute back $y = \log_3 x$ to find the values of $x$. For $y = \frac{2}{3}$: $\log_3 x = \frac{2}{3}$ $x = 3^{2/3}$ This solution $x = 3^{2/3} = \sqrt[3]{9}$ is positive, so it is valid.

For $y = 2$: $\log_3 x = 2$ $x = 3^2$ $x = 9$ This solution $x = 9$ is positive, so it is valid.

The solutions to the equation are $x = 3^{2/3}$ and $x = 9$.