Question

select the correct answer.
what is the solution to this equation?
\log_{4}(x^{2}+1)=\log_{4}(-2x)
a. \\ x = 2
b. \\ x = 1
c. \\ x = -1
d. \\ no solution

Explanation:

Step1: Set arguments equal (log property)

If $\log_b(a) = \log_b(c)$, then $a = c$ (for $b>0, b
eq1, a>0, c>0$).
So $x^2 + 1 = -2x$

Step2: Rearrange to quadratic equation

$x^2 + 2x + 1 = 0$

Step3: Factor the quadratic

$(x+1)^2 = 0$
Solve for $x$: $x = -1$

Step4: Verify domain validity

Check if arguments are positive:
For $x=-1$: $x^2+1 = (-1)^2+1=2>0$, $-2x=-2(-1)=2>0$. Both are valid.

Answer:

C. $x = -1$