Sure, let's solve the equation [tex]\(\ln (x+5) = \ln (x-1) - \ln (x+1)\)[/tex].
### Step 1: Combine the Logarithms on the Right Side
First, we use the logarithm property [tex]\(\ln a - \ln b = \ln \left(\frac{a}{b}\)[/tex]\):
[tex]\[
\ln (x-1) - \ln (x+1) = \ln \left( \frac{x-1}{x+1} \right)
\][/tex]
So the equation now looks like this:
[tex]\[
\ln (x+5) = \ln \left( \frac{x-1}{x+1} \right)
\][/tex]
### Step 2: Use the Property of Logarithms
Next, we use the fact that if [tex]\(\ln a = \ln b\)[/tex], then [tex]\(a = b\)[/tex]. Applying this property:
[tex]\[
x+5 = \frac{x-1}{x+1}
\][/tex]
### Step 3: Solve the Equation
To eliminate the fraction, multiply both sides by [tex]\(x+1\)[/tex]:
[tex]\[
(x+5)(x+1) = x-1
\][/tex]
Now, expand and simplify:
[tex]\[
x^2 + 6x + 5 = x - 1
\][/tex]
Move all terms to one side to form a quadratic equation:
[tex]\[
x^2 + 6x + 5 - x + 1 = 0
\][/tex]
[tex]\[
x^2 + 5x + 6 = 0
\][/tex]
### Step 4: Factor the Quadratic Equation
To find the roots, we factor the quadratic equation:
[tex]\[
x^2 + 5x + 6 = (x + 2)(x + 3) = 0
\][/tex]
Set each factor equal to zero:
[tex]\[
x + 2 = 0 \quad \text{or} \quad x + 3 = 0
\][/tex]
Thus, the solutions are:
[tex]\[
x = -2 \quad \text{or} \quad x = -3
\][/tex]
### Step 5: Verify the Solutions
Finally, we need to check whether these solutions make the original logarithmic expressions valid (since logarithms are only defined for positive arguments).
For [tex]\(x = -2\)[/tex]:
[tex]\[
x + 5 = -2 + 5 = 3 \quad (\ln(3) \text{ is valid})
\][/tex]
[tex]\[
x - 1 = -2 - 1 = -3 \quad (\ln(-3) \text{ is not valid})
\][/tex]
For [tex]\(x = -3\)[/tex]:
[tex]\[
x + 5 = -3 + 5 = 2 \quad (\ln(2) \text{ is valid})
\][/tex]
[tex]\[
x - 1 = -3 - 1 = -4 \quad (\ln(-4) \text{ is not valid})
\][/tex]
Since both values make the logarithmic expressions invalid, there are no valid solutions to this equation.
Thus, the original equation [tex]\(\ln (x+5) = \ln (x-1) - \ln (x+1)\)[/tex] has no solutions.
