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Question

(A) Given the equation: $e^{5(\ln x)^2+3} = x^8$ Taking the natural logarithm $(\ln)$ on both sides: $\ln(e^{5(\ln x)^2+3}) = \ln(x^8)$ $5(\ln x)^2 +

Vedclass · Accepted Answer

$e^{8/5}$

Vedclass · Answer

(A) Given the equation: $e^{5(\ln x)^2+3} = x^8$ Taking the natural logarithm $(\ln)$ on both sides: $\ln(e^{5(\ln x)^2+3}) = \ln(x^8)$ $5(\ln x)^2 + 3 = 8 \ln x$ Let $t = \ln x$. The equation becomes: $5t^2 - 8t + 3 = 0$ This is a quadratic equation in $t$. Let the roots be $t_1$ and $t_2$. By the properties of quadratic equations,the sum of the roots is $t_1 + t_2 = -(-8)/5 = 8/5$. Since $t = \ln x$,we have $\ln x_1 + \ln x_2 = 8/5$. Using the property $\ln x_1 + \ln x_2 = \ln(x_1 x_2)$,we get: $\ln(x_1 x_2) = 8/5$ Therefore,the product of the solutions is $x_1 x_2 = e^{8/5}$.

The product of all solutions of the equation $e^{5(\ln x)^2+3} = x^8$,where $x > 0$,is:

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