If $\log x-5\log 3=-2$, then $x$ equals
$0.8$
(Logs are base 10.) Move the term to the RHS: \[ \log x=-2+5\log 3=\log\!\big(10^{-2}\big)+\log\!\big(3^5\big)=\log\!\Big(\frac{3^5}{100}\Big). \] Hence \(x=\dfrac{3^5}{100}=\dfrac{243}{100}=2.43.\)
The product of all solutions of the equation \(e^{5(\log_e x)^2 + 3 = x^8, x > 0}\) , is :