Solve for \( x \) : \( \log (x)+\log (x+1)=4 \) \( x=\square \) You may enter the exact value or round to 4 decimal places. Question Help: Video

To solve the equation \( \log (x) + \log (x+1) = 4 \), we can use the property of logarithms that states \( \log(a) + \log(b) = \log(ab) \). Thus, we can rewrite the equation as: \[ \log(x(x + 1)) = 4 \] Exponentiating both sides to eliminate the logarithm gives us: \[ x(x + 1) = 10^4 \] This simplifies to: \[ x^2 + x - 10000 = 0 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, \) and \( c = -10000 \): \[ b^2 - 4ac = 1^2 - 4(1)(-10000) = 1 + 40000 = 40001 \] Thus, the solutions for \( x \) are: \[ x = \frac{-1 \pm \sqrt{40001}}{2} \] Evaluating \(\sqrt{40001} \approx 200.0025\) gives us: \[ x \approx \frac{-1 + 200.0025}{2} \approx 99.50125 \] \[ x \approx \frac{-1 - 200.0025}{2} \text{ (not relevant, as x must be positive)} \] After rounding to four decimal places, we have: \( x \approx 99.5013 \)