Question

which graph shows the solution to the equation \\(\log_{2}(3x - 1) = 2\\)?

Explanation:

Step1: Solve the logarithmic equation

To solve \(\log_{2}(3x - 1)=2\), we use the definition of a logarithm. If \(\log_{b}(a)=c\), then \(a = b^{c}\). So, applying this to our equation, we get:
\(3x - 1=2^{2}\)

Step2: Simplify the right - hand side

Calculate \(2^{2}\), which is equal to 4. So the equation becomes:
\(3x - 1 = 4\)

Step3: Solve for x

Add 1 to both sides of the equation:
\(3x=4 + 1\)
\(3x=5\)
Then divide both sides by 3:
\(x=\frac{5}{3}\approx1.67\)

Now, we also know that the equation \(\log_{2}(3x - 1)=2\) can be thought of as finding the intersection of the function \(y = \log_{2}(3x - 1)\) and the horizontal line \(y = 2\).

First, let's analyze the domain of the logarithmic function \(y=\log_{2}(3x - 1)\). For a logarithmic function \(\log_{b}(u)\), \(u>0\). So \(3x-1>0\), which gives \(x>\frac{1}{3}\).

Now, let's find the value of \(x\) when \(y = 2\) (from the line \(y = 2\)) and see which graph has the intersection at \(x=\frac{5}{3}\approx1.67\).

Looking at the first graph, the x - axis has markings up to 3, and the second graph has x - axis markings up to 6. But when we solve \(x=\frac{5}{3}\approx1.67\), we can check the horizontal line \(y = 2\) (since the equation is \(\log_{2}(3x - 1)=2\), so \(y = 2\) for the horizontal line) and the logarithmic curve \(y=\log_{2}(3x - 1)\).

The key is that when \(y = 2\), we found \(x=\frac{5}{3}\approx1.67\). Now, let's check the horizontal line: the equation is \(\log_{2}(3x - 1)=2\), so the horizontal line should be \(y = 2\).

In the first graph, the horizontal line is \(y = 3\) (wait, no, looking back, the first graph's horizontal line is at \(y = 3\)? Wait, no, the first graph's orange line: looking at the y - axis, the first graph has y - axis with 2, 4? Wait, no, let's re - examine.

Wait, the equation is \(\log_{2}(3x - 1)=2\), so we are looking for the intersection of \(y=\log_{2}(3x - 1)\) and \(y = 2\).

Let's recast the function \(y=\log_{2}(3x - 1)\). Let's find the x - intercept: set \(y = 0\), then \(\log_{2}(3x - 1)=0\), so \(3x - 1=1\), \(3x=2\), \(x=\frac{2}{3}\approx0.67\).

Now, when \(y = 2\), \(x=\frac{5}{3}\approx1.67\).

Looking at the two graphs:

The first graph: the horizontal line (orange) is at \(y = 3\)? No, wait the first graph's y - axis: the top orange line is at \(y = 3\)? Wait, no, the first graph's y - axis has 2, 4? Wait, the first graph's orange line is horizontal, let's check the y - coordinate. The equation is \(\log_{2}(3x - 1)=2\), so the horizontal line should be \(y = 2\).

Wait, maybe I misread the graphs. Let's start over.

The equation is \(\log_{2}(3x - 1)=2\). To solve it:

1. Convert logarithmic to exponential: \(3x-1 = 2^{2}=4\)
2. Then \(3x=5\), \(x=\frac{5}{3}\approx1.67\)

Now, the function \(y = \log_{2}(3x - 1)\) is a logarithmic function. The domain is \(3x-1>0\Rightarrow x>\frac{1}{3}\approx0.33\).

The horizontal line is \(y = 2\) (since the equation is \(\log_{2}(3x - 1)=2\), so we set \(y = 2\)).

Now, let's look at the two graphs:

- First graph: The horizontal line (orange) is at \(y = 3\)? No, the first graph's y - axis: the orange line is at \(y = 3\)? Wait, no, the first graph's y - axis has markings: the first graph's y - axis, the orange line is at \(y = 3\)? Wait, no, the user's graph: first graph, y - axis: 2, 4? Wait, the first graph: the orange line is horizontal, and the blue curve is a log curve.

Wait, the second graph: the horizontal line is at \(y = 2\), and the blue curve. Let's check the x - value when \(y = 2\).

For the second graph, when \(y = 2\), the x - value of intersection: let's see, the blue curve is \(y=\log_{2}(3x - 1)\). When \(y = 2\), \(x=\frac{5}{3}\approx1.67\), which is between 0 and 2 (since the second graph's x - axis has 2, 6). Wait, the first graph's x - axis has 1, 2, 3.

Wait, maybe the key is that the equation \(\log_{2}(3x - 1)=2\) is equivalent to finding where \(y=\log_{2}(3x - 1)\) and \(y = 2\) intersect. So we solve for \(x\) and then see which graph has that intersection.

We found \(x=\frac{5}{3}\approx1.67\). Now, let's check the horizontal line: the equation is \(\log_{2}(3x - 1)=2\), so the horizontal line is \(y = 2\).

In the first graph, the horizontal line is at \(y = 3\) (no, the first graph's orange line: looking at the y - axis, the first graph's y - axis has 2, 4? Wait, the first graph's y - axis: the top orange line is at \(y = 3\)? Wait, no, the user's first graph: the orange line is horizontal, and the blue curve is a log curve. The second graph: orange line at \(y = 2\), blue curve.

Wait, the equation is \(\log_{2}(3x - 1)=2\), so \(y = 2\) for the horizontal line. So the horizontal line should be \(y = 2\). So the second graph has the horizontal line at \(y = 2\), and the first graph's horizontal line is at a higher y - value (maybe \(y = 3\) or something else).

Also, when we solve \(x=\frac{5}{3}\approx1.67\), which is between 1 and 2. So the intersection point should be at \(x\approx1.67\) and \(y = 2\).

So the graph where the horizontal line \(y = 2\) (since the equation is \(\log_{2}(3x - 1)=2\)) intersects the logarithmic curve \(y=\log_{2}(3x - 1)\) at \(x=\frac{5}{3}\approx1.67\) is the correct one.

Assuming that the second graph (the one with the horizontal line at \(y = 2\) and the x - axis with 2, 6) has the intersection at \(x\approx1.67\) (since \(\frac{5}{3}\approx1.67\) is between 1 and 2), and the first graph's horizontal line is at a different y - value (maybe \(y = 3\)) and the intersection x - value is different.

So the correct graph is the one where the horizontal line is \(y = 2\) (since the equation is \(\log_{2}(3x - 1)=2\)) and the logarithmic curve intersects it at \(x=\frac{5}{3}\approx1.67\).

Answer:

The graph (the second one, with the horizontal line \(y = 2\) and the x - axis marked with 2, 6) shows the solution. (Assuming the second graph in the options has the horizontal line \(y = 2\) and the logarithmic curve intersecting it at \(x=\frac{5}{3}\approx1.67\))