Question

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

Explanation:

To solve the equation \(\log(2x + 1)=3x - 2\) graphically, we need to set each side of the equation equal to a separate function \(y_1\) and \(y_2\). The solution to the equation will be the \(x\)-coordinate of the intersection point of the graphs of \(y_1\) and \(y_2\).

Step 1: Analyze the left - hand side (LHS) of the equation

The left - hand side of the equation \(\log(2x + 1)=3x - 2\) is \(\log(2x + 1)\). We can set \(y_1=\log(2x + 1)\).

Step 2: Analyze the right - hand side (RHS) of the equation

The right - hand side of the equation \(\log(2x + 1)=3x - 2\) is \(3x-2\). We can set \(y_2 = 3x - 2\).

Now let's analyze the other options:

• Option 1: \(y_1 = 3x\) and \(y_2=2x\) have nothing to do with the original equation \(\log(2x + 1)=3x - 2\), so this option is incorrect.
• Option 3: The expression \(y_1=\log2x + 1\) is not the same as \(\log(2x + 1)\) (the parentheses are important for the logarithm's argument). The argument of the logarithm in the original equation is \(2x + 1\), and in \(y_1=\log2x+1\), the argument of the logarithm is \(2x\) (assuming the base - 10 logarithm, and the expression is mis - written), so this option is incorrect.
• Option 4: \(y_1=\log(2x + 1 + 2)=\log(2x+3)\) and \(y_2 = 3x\) do not match the original equation, so this option is incorrect.

Answer:

B. \(y_1=\log(2x + 1),y_2 = 3x-2\)