Need to solve for q.
f=.2083(100/c)^1.852*q^1.852/dh^4.8655*5280/100
This is an equation of the form:
$\displaystyle f = \frac {Aq^{1.852}} B$
where A and B are constants: $\displaystyle A = 0.2083 (\frac {100} C ) ^{1.852}$ and $\displaystyle B=dh^{4.8655} (\frac {5280} {100})$. So you can rearrange as follows:
$\displaystyle q^{1.852} = \frac {fB} A$
$\displaystyle 1.852 \log(q) = \log ( \frac {fB} A )$
$\displaystyle \log (q) = \frac 1 {1.852} \log ( \frac {fB} A ) $
$\displaystyle q = e^{ [ \frac 1 {1.852} \log(fB/A) ]} = (\frac {fB} A )^\frac 1 {1.852}$