Certainly! Let’s solve the equation step by step:
[tex]\[ \log(x-1) + \log(x+1) = 2\log(x+2) \][/tex]
### Step 1: Apply the properties of logarithms
Using the property [tex]\(\log a + \log b = \log(ab)\)[/tex] on the left-hand side:
[tex]\[ \log\left((x-1)(x+1)\right) = 2\log(x+2) \][/tex]
### Step 2: Simplify the equation
Expand the left-hand side:
[tex]\[ \log(x^2 - 1) = 2\log(x+2) \][/tex]
Use the property [tex]\(a\log b = \log(b^a)\)[/tex] on the right-hand side:
[tex]\[ \log(x^2 - 1) = \log((x+2)^2) \][/tex]
### Step 3: Remove the logarithms using the property that if [tex]\(\log a = \log b\)[/tex], then [tex]\(a = b\)[/tex]:
[tex]\[ x^2 - 1 = (x+2)^2 \][/tex]
### Step 4: Expand and simplify the quadratic equation
[tex]\[ x^2 - 1 = x^2 + 4x + 4 \][/tex]
Subtract [tex]\(x^2\)[/tex] from both sides:
[tex]\[ -1 = 4x + 4 \][/tex]
### Step 5: Solve for [tex]\(x\)[/tex]
Subtract 4 from both sides:
[tex]\[ -5 = 4x \][/tex]
Divide by 4:
[tex]\[ x = -\frac{5}{4} \][/tex]
### Step 6: Verify the result
We need to ensure that the solution satisfies the original domain requirements, where the arguments of the logarithms must be positive:
- [tex]\(x - 1 > 0 \implies x > 1\)[/tex]
- [tex]\(x + 1 > 0 \implies x > -1\)[/tex]
- [tex]\(x + 2 > 0 \implies x > -2\)[/tex]
From these conditions, [tex]\(x > 1\)[/tex] is the most restrictive.
The solution [tex]\(x = -\frac{5}{4}\)[/tex] does NOT satisfy this condition.
### Conclude there are no real solutions
Therefore, there are no real values of [tex]\(x\)[/tex] that satisfy the given equation.
