\( \equiv \) L) Listen Solve \( \log _{4}(-x)+\log _{4}(x+10)=2 \). Checkifor extraneous solutions. \( x=\square \) and \( x=\square \)

To solve the equation \( \log _{4}(-x)+\log _{4}(x+10)=2 \), start by combining the logarithms: \[ \log_{4}((-x)(x+10)) = 2 \] This means that: \[ (-x)(x+10) = 4^2 = 16 \] So, we can rewrite the equation: \[ -x^2 - 10x - 16 = 0 \] Rearranging gives: \[ x^2 + 10x + 16 = 0 \] Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} \] \[ x = \frac{-10 \pm \sqrt{100 - 64}}{2} \] \[ x = \frac{-10 \pm \sqrt{36}}{2} \] \[ x = \frac{-10 \pm 6}{2} \] This gives us two possible solutions: 1. \( x = \frac{-4}{2} = -2 \) 2. \( x = \frac{-16}{2} = -8 \) Now let's check for extraneous solutions. Substitute \( x = -2 \): \[ \log_4(-(-2)) + \log_4(-2 + 10) = \log_4(2) + \log_4(8) = \log_4(2 \times 8) = \log_4(16) = 2 \] This is valid! Now check \( x = -8 \): \[ \log_4(-(-8)) + \log_4(-8 + 10) = \log_4(8) + \log_4(2) \] This becomes: \[ \log_4(8 \times 2) = \log_4(16) = 2 \] This is also valid! Thus, both solutions are correct: \( x = -2 \) and \( x = -8 \). To ensure clarity while dealing with logarithmic equations, knowing how to check for the domain of the logarithmic function is crucial. Logarithms are defined only for positive numbers, so you must always verify that the arguments of the logs remain positive throughout the operations. Understanding the properties of logarithmic identities can make solving logarithmic equations much simpler. Familiarize yourself with key transformations, such as the change of base formula and how to condense or expand logarithmic expressions. These tools can come in handy and save you time solving similar problems in the future!