Approximate the solution to the equation \( 20=125 e^{x} \) by locating the intersection point of \( y=125 e^{x} \) and \( y=20 \). (1 point) \( x \approx 20 \) \( x \approx 125 \) \( x \approx-1.8 \) \( x \approx 1.8 \)

To approximate the solution to the equation \( 20 = 125 e^{x} \), we can rewrite it in the form \( y = 125 e^{x} \) and then determine where this curve intersects the horizontal line \( y = 20 \). Starting with \( y = 125 e^{x} \), we see that as \( x \) decreases, \( e^{x} \) approaches zero, and as \( x \) increases, \( e^{x} \) grows rapidly. By substituting various values for \( x \), we can estimate where the intersection occurs. Testing \( x = -1.8 \) gives approximately \( y \approx 125 e^{-1.8} \), which is roughly \( 20 \). Confirming through calculation will show that this value is indeed the intersection, leading us to conclude that the correct option is \( x \approx -1.8 \).