Log then4(x squre plus x)-log then4(x plus 1)2 what is x?

Answer 1

To solve the equation (\log_{4}(x^2 + x) - \log_{4}((x + 1)^2) = 0), we can use the properties of logarithms. This simplifies to (\log_{4}\left(\frac{x^2 + x}{(x + 1)^2}\right) = 0), which implies (\frac{x^2 + x}{(x + 1)^2} = 1). Solving this gives (x^2 + x = (x + 1)^2), leading to (x^2 + x = x^2 + 2x + 1), resulting in (x = -1). However, substituting (x = -1) back into the original logarithmic expression yields an undefined result, so there are no valid solutions for (x).