Answer :
To solve the equation [tex]\log_5(1-x) = -\log_5(x-4)[/tex], we need to use the properties of logarithms and some algebra.
Start by using the property of negative logarithms:
[tex]-\log_5(x-4) = \log_5\left(\frac{1}{x-4}\right)[/tex]
So, the original equation becomes:
[tex]\log_5(1-x) = \log_5\left(\frac{1}{x-4}\right)[/tex]
Since the logarithms on both sides have the same base, their arguments must be equal:
[tex]1-x = \frac{1}{x-4}[/tex]
Clear the fraction by multiplying both sides by [tex]x-4[/tex]:
[tex](1-x)(x-4) = 1[/tex]
Expand the left side:
[tex]1\cdot x - 1\cdot 4 - x\cdot x + x\cdot 4 = 1[/tex]
[tex]x - 4 - x^2 + 4x = 1[/tex]Simplify the expression:
[tex]-x^2 + 5x - 4 = 1[/tex]
Rearrange it into a standard quadratic equation:
[tex]-x^2 + 5x - 4 - 1 = 0[/tex]
[tex]-x^2 + 5x - 5 = 0[/tex]Multiply the whole equation by -1 to make it easier to solve:
[tex]x^2 - 5x + 5 = 0[/tex]
Use the quadratic formula to solve for [tex]x[/tex]:
The quadratic formula is given by [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]. Here, [tex]a = 1[/tex], [tex]b = -5[/tex], and [tex]c = 5[/tex].
[tex]x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4\cdot 1 \cdot 5}}{2\cdot 1}[/tex]
[tex]x = \frac{5 \pm \sqrt{25 - 20}}{2}[/tex]
[tex]x = \frac{5 \pm \sqrt{5}}{2}[/tex]Thus, the solutions for [tex]x[/tex] are:
- [tex]x = \frac{5 + \sqrt{5}}{2}[/tex]
- [tex]x = \frac{5 - \sqrt{5}}{2}[/tex]
Check for valid solutions:
Both solutions need to satisfy the conditions [tex]1-x > 0[/tex] and [tex]x-4 > 0[/tex].
For [tex]x = \frac{5 + \sqrt{5}}{2}[/tex], these conditions are not satisfied as [tex]1-x[/tex] is negative.
For [tex]x = \frac{5 - \sqrt{5}}{2}[/tex], neither condition is satisfied too.
Conclusion:
Therefore, there might have been an oversight in how the conditions were checked, but it's critical to confirm these solutions against the domain restrictions. Given the choices however, the arithmetic simplifies to the choice given in (c) [tex]\frac{-5 \pm \sqrt{5}}{2}[/tex], but the domain check shows neither could potentially work given the logarithmic constraints.
The correct option is (d) None of these since [tex]x[/tex] must ensure the arguments of the logarithms remain positive, and neither satisfies these conditions based on initial computations.