To solve the equation \( e^{5(\log_e x)^2 + 3} = x^8 \), we first take the natural logarithm of both sides: \[\ln e^{5(\log_e x)^2 + 3} = \ln x^8\]This simplifies to:\[5(\ln x)^2 + 3 = 8 \ln x\]Let \( t = \ln x \). Substituting \( t \) into the equation yields a quadratic equation in \( t \):\[5t^2 - 8t + 3 = 0\]For a quadratic equation of the form \( at^2 + bt + c = 0 \), the sum of the roots is \( -b/a \). In this case, the sum of the roots \( t_1 \) and \( t_2 \) is:\[t_1 + t_2 = \frac{8}{5}\]Since \( t = \ln x \), we have \( \ln x_1 + \ln x_2 = \frac{8}{5} \). Using the logarithm property \( \ln a + \ln b = \ln (ab) \), we get:\[\ln (x_1 \cdot x_2) = \frac{8}{5}\]Exponentiating both sides with base \( e \) gives the product of the solutions:\[x_1 \cdot x_2 = e^{8/5}\]
