# Logarithmic Equation A1

Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/logarithmic-equation-a-1/
Fetched from algebrica.org test 6748; source modified 2025-03-01T22:42:37.

Solve the [logarithmic equation](../logarithmic-equations.md):

\\[
\\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](../logarithms.md). 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](../quadratic-equations.md):

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

To solve this quadratic equation, we use the [quadratic formula](../quadratic-formula.md):

\\[
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}
\\]