Solve for \( x \) \( \log _{5} x+\log _{5}(x+1)=9 \) \( x=\square \)

To solve the equation \( \log _{5} x + \log _{5}(x+1) = 9 \), we can use the property of logarithms that states \( \log_b a + \log_b c = \log_b (ac) \). Applying this property, we rewrite the left-hand side: \[ \log_{5}(x(x+1)) = 9. \] This implies: \[ x(x + 1) = 5^9. \] Calculating \( 5^9 \): \[ 5^9 = 1953125. \] So the equation now reads: \[ x^2 + x - 1953125 = 0. \] We can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1, b=1, c=-1953125 \): \[ b^2 - 4ac = 1^2 - 4(1)(-1953125) = 1 + 7812500 = 7812501. \] Taking the square root: \[ \sqrt{7812501} \approx 2795.50. \] Now, substituting back into the quadratic formula: \[ x = \frac{-1 \pm 2795.50}{2}. \] Calculating the two possible values for \( x \): 1. \( x = \frac{2794.50}{2} \approx 1397.25 \) 2. \( x = \frac{-2796.50}{2} \) (which is negative and not valid since \( x \) must be positive) Thus, the solution is: \[ x \approx 1397.25. \] Finally: \[ x \approx 1397.25 \quad \text{(keep rounding as necessary)}. \]