Given equation

$e^{5(\ln x)^2+3}=x^8, \qquad x>0$

Since $x>0$, let $t=\ln x$ so that $x=e^t.$

Rewrite the equation in terms of $t$

We know $x^8=(e^t)^8=e^{8t}.$ Therefore, $e^{5t^2+3}=e^{8t}.$

Since exponential function is one-one, $5t^2+3=8t.$

Solve the quadratic

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

Factorizing, $5t^2-5t-3t+3=0$ $5t(t-1)-3(t-1)=0$ $(5t-3)(t-1)=0$

So, $t=1 \quad \text{or} \quad t=\frac35.$

Find corresponding values of $x$

Since $x=e^t$, $x=e^1=e$ and $x=e^{3/5}.$

Product of all solutions

$e\cdot e^{3/5}=e^{1+3/5}=e^{8/5}.$

Compare with options

The product is $\boxed{e^{8/5}}$ which is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer also gives D.