JEE Main 2025 — Logarithm Question with Solution

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:

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