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 \(x=\dfrac{3^5}{100}=\dfrac{243}{100}=2.43.\)
Consider two distinct positive numbers \( m, n \) with \( m > n \). Let \[ x = n^{\log_n m}, \quad y = m^{\log_m n}. \] The relation between \( x \) and \( y \) is -
If \[ \log_{p^{1/2}} y \times \log_{y^{1/2}} p = 16, \] then find the value of the given expression.