Logarithmic Equation A1

Solve the logarithmic equation:

\[ \log_2 (x+2) + \log_2 (x-1) = \log_2 (6) \]

The first step is to determine the domain of the equation. We have: \[ \log_af(x) = \begin{cases} a > 0 \\[0.6em] a \neq 1 \\[0.6em] f(x) > 0 \\ \end{cases} \]

This gives us the following conditions:

• The argument (x + 2) must be greater than zero, so (x > -2).
• The argument (x - 1) must be greater than zero, so (x > 1).

Since both conditions must hold, the overall domain is (x > 1).

Next, we apply the logarithmic product rule. This rule states that:

\[ \log_a(m) + \log_a(n) = \log_a(m \cdot n) \]

Using this rule, we can combine the two logarithms on the left-hand side into one:

\[ \log_2\big[(x + 2)(x - 1)\big] = \log_2 (6) \]

Since the logarithms on both sides have the same base, their arguments must be equal. Therefore, we set:

\[ (x + 2)(x - 1) = 6 \]

We now solve the resulting equation. First, we expand the product:

\[ x^2 + x - 2 = 6 \]

Then, we bring all terms to one side to form a quadratic equation:

\[ x^2 + x - 8 = 0 \]

To solve this quadratic equation, we use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

with (a = 1), (b = 1), and (c = -8). The discriminant is:

\[ \Delta = 1^2 - 4(1)(-8) = 1 + 32 = 33 \]

Thus, the solutions are:

\[ x = \frac{-1 \pm \sqrt{33}}{2} \]

The two potential solutions are:

\[ x_1 = \frac{-1 + \sqrt{33}}{2} \quad \text{and} \quad x_2 = \frac{-1 - \sqrt{33}}{2} \]

We now test these solutions against the domain \(x > 1\). Approximating \(\sqrt{33} \approx 5.744\), we find:

\[ x_1 \approx \frac{-1 + 5.744}{2} \approx 2.372 \quad (\text{valid since } 2.372 > 1) \] \[ x_2 \approx \frac{-1 - 5.744}{2} \approx -3.372 \quad (\text{invalid since } -3.372 \not> 1) \]

Thus, only \(x_1\) is an acceptable solution.