Question

which system of equations could be graphed to solve the equation?\\(\log(2x + 1) = 3x - 2\\)\\(y_1 = 3x, y_2 = 2x\\)\\(y_1 = \log 2x + 1, y_2 = 3x - 2\\)\\(y_1 = \log(2x + 1), y_2 = 3x - 2\\)\\(y_1 = \log(2x + 1 + 2), y_2 = 3x\\)

Explanation:

Step1: Recall the method to solve equations by graphing

To solve an equation \( f(x) = g(x) \) by graphing, we can graph the two functions \( y_1 = f(x) \) and \( y_2 = g(x) \), and the \( x \)-coordinates of the intersection points are the solutions of the equation.

Step2: Analyze the given equation

The given equation is \( \log(2x + 1)=3x - 2 \). So we can let \( y_1=\log(2x + 1) \) (the left - hand side of the equation) and \( y_2 = 3x-2 \) (the right - hand side of the equation).

Step3: Check the options

• For the first option \( y_1 = 3x,y_2 = 2x \), these two functions are not related to the given equation \( \log(2x + 1)=3x - 2 \), so this option is incorrect.
• For the second option \( y_1=\log(2x)+1,y_2 = 3x - 2 \), the left - hand side function \( \log(2x)+1\) is not the same as \( \log(2x + 1) \) in the given equation, so this option is incorrect.
• For the third option \( y_1=\log(2x + 1),y_2 = 3x - 2 \), this option correctly represents the left - hand side and the right - hand side of the equation \( \log(2x + 1)=3x - 2 \) as two separate functions for graphing.
• For the fourth option \( y_1=\log(2x + 1 + 2),y_2 = 3x \), the left - hand side function \( \log(2x+3) \) and the right - hand side function \( 3x \) are not related to the given equation \( \log(2x + 1)=3x - 2 \), so this option is incorrect.

Answer:

\( y_1=\log(2x + 1),y_2 = 3x - 2 \) (the third option)