QJEEM25Ma87d9: JEE Main 2025 Mathematics - Class 12 JEEM25Ma87d9 Solution: The product of all solutions of the equation , is :

We begin with the equation:

e^{5(\log_e x)^2 + 3} = x^8, \; x > 0.

Equating the exponents, we have:

$5(\log_e x)^2 + 3 = \log_e x^8 = 8 \log_e x.$

Let $t = \log_e x$ . Substituting, the equation becomes:

$5t^2 + 3 = 8t.$

Rewriting this as a quadratic equation:

$5t^2 - 8t + 3 = 0.$

Factoring the quadratic:

$5t^2 - 5t - 3t + 3 = 0,$

$(5t - 3)(t - 1) = 0.$

Thus, the solutions for $t$ are:

$t = 1$ implies $\log_e x = 1$ , giving $x = e$ .

$t = \frac{3}{5}$ implies $\log_e x = \frac{3}{5}$ , giving $x = e^{\frac{3}{5}}$ .

The product of all solutions is:

$e^1 \times e^{\frac{3}{5}} = e^{1 + \frac{3}{5}} = e^{\frac{8}{5}}.$