Logarithmic Equation A1

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\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:

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.

The solution to the equation is:

x = \frac{-1 + \sqrt{33}}{2}