Solve for x log base 4 of x^2+1 = log base 4 of -2x

$\log_{4} (x^{2} + 1) = \log_{4} (- 2 x)$

Step 1

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} + 1 = - 2 x$

Step 2

Solve for $x$ .

Add $2 x$ to both sides of the equation.

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

Factor using the perfect square rule.

Rearrange terms.

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

Rewrite $1$ as $1^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x + 1)}^{2} = 0$

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$