To solve the equation [tex]\(\ln (-x + 1) - \ln (3x + 5) = \ln (-6x + 1)\)[/tex], we start by using logarithmic properties to combine the logarithmic expressions.
First, recall that [tex]\(\ln a - \ln b = \ln \left(\frac{a}{b}\right)\)[/tex]. Applying this property, we get:
[tex]\[
\ln \left(\frac{-x + 1}{3x + 5}\right) = \ln (-6x + 1)
\][/tex]
For the logarithms to be equal, their arguments must be equal. Thus, we remove the logarithms and set the arguments equal to each other:
[tex]\[
\frac{-x + 1}{3x + 5} = -6x + 1
\][/tex]
Next, we clear the fraction by multiplying both sides by [tex]\(3x + 5\)[/tex]:
[tex]\[
-x + 1 = (-6x + 1)(3x + 5)
\][/tex]
Now we expand the right-hand side:
[tex]\[
-x + 1 = (-6x + 1)(3x + 5)\\
-x + 1 = -18x^2 - 30x + 3x + 5\\
-x + 1 = -18x^2 - 27x + 5
\][/tex]
Rearrange the equation to set it to zero:
[tex]\[
-x + 1 = -18x^2 - 27x + 5\\
0 = -18x^2 - 27x + 5 + x - 1\\
0 = -18x^2 - 26x + 4
\][/tex]
Now we have a quadratic equation:
[tex]\[
-18x^2 - 26x + 4 = 0
\][/tex]
We can solve this quadratic equation using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = -18\)[/tex], [tex]\(b = -26\)[/tex], and [tex]\(c = 4\)[/tex]:
[tex]\[
x = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(-18)(4)}}{2(-18)}\\
x = \frac{26 \pm \sqrt{676 + 288}}{-36}\\
x = \frac{26 \pm \sqrt{964}}{-36}\\
x = \frac{26 \pm 31.05}{-36}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{26 + 31.05}{-36} = \frac{57.05}{-36} \approx -1.58
\][/tex]
[tex]\[
x = \frac{26 - 31.05}{-36} = \frac{-5.05}{-36} \approx 0.14
\][/tex]
Finally, we must verify which of these solutions are valid in the original logarithmic equation constraints, ensuring the arguments of the logarithms are positive.
- For [tex]\(x \approx -1.58\)[/tex]:
[tex]\[
-x + 1 = -(-1.58) + 1 = 2.58 \quad (\text{positive})\\
3x + 5 = 3(-1.58) + 5 = -4.74 + 5 = 0.26 \quad (\text{positive})\\
-6x + 1 = -6(-1.58) + 1 = 10.48 \quad (\text{positive})
\][/tex]
[tex]\(x = -1.58\)[/tex] satisfies the constraints.
- For [tex]\(x \approx 0.14\)[/tex]:
[tex]\[
-x + 1 = -0.14 + 1 = 0.86 \quad (\text{positive})\\
3x + 5 = 3(0.14) + 5 = 0.42 + 5 = 5.42 \quad (\text{positive})\\
-6x + 1 = -6(0.14) + 1 = -0.84 + 1 = 0.16 \quad (\text{positive})
\][/tex]
[tex]\(x = 0.14\)[/tex] also satisfies the constraints.
The correct pair of solutions for the given quadratic equation is:
-1.58 or 0.14
