Solve for \( x \) : \( \log _{5} x+\log _{5}(x+3)=3 \) \( x=\square \) Question Help: Video
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To solve the equation \( \log _{5} x + \log _{5}(x+3) = 3 \), we can combine the logs on the left side using the property of logarithms that states \( \log_a b + \log_a c = \log_a (b \cdot c) \). This gives us: \[ \log_{5} (x(x + 3)) = 3 \] Now, we can rewrite the equation in exponential form: \[ x(x + 3) = 5^3 \] Calculating \( 5^3 \) gives us 125, so we have: \[ x(x + 3) = 125 \] This simplifies to a quadratic equation: \[ x^2 + 3x - 125 = 0 \] To solve for \( x \), we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 3, c = -125 \). Plugging in these values: \[ b^2 - 4ac = 3^2 - 4(1)(-125) = 9 + 500 = 509 \] Now substituting into the quadratic formula: \[ x = \frac{-3 \pm \sqrt{509}}{2} \] Calculating \( \sqrt{509} \) gives approximately 22.56, so: \[ x \approx \frac{-3 + 22.56}{2} \approx \frac{19.56}{2} \approx 9.78 \] Since \( x \) must be positive and \( \frac{-3 - \sqrt{509}}{2} \) would yield a negative value, we only consider the solution: \( x \approx 9.78 \) which we can round to \( x \approx 9.8 \). So the final answer is: \( x = 9.8 \)
