Solve the equation $2 x \lg x+x-1=0$ in the set of real numbers. (Proposed by Éva Gyanó, Budapest)

Solve the equation $2 x \lg x+x-1=0$ in the set of real numbers. (Proposed by Éva Gyanó, Budapest)
Solution. Let's rearrange the equation:

$$
2 \lg x=\frac{1}{x}-1
$$

then plot the functions on the left and right side of the equation in the same coordinate system. From the graph, we can read that both functions take the value 0 at \( x=1 \). Substitution can verify that indeed \( 2 \lg 1=0 \) and \( \frac{1}{1}-1=0 \).

![](https://cdn.mathpix.com/cropped/2024_05_02_082aee7b1c3c36019951g-1.jpg?height=374&width=593&top_left_y=469&top_left_x=755)

Since \( 2 \lg x \) is strictly increasing and \( \frac{1}{x}-1 \) is strictly decreasing for \( x>0 \), they can only be equal at \( x=1 \). The only solution to the equation is \( x=1 \).