Calculating Fluoride in toothpaste Question
  A linear regression line was obtained; $$\log(\ce{[NaF]})~(\mathrm{mol~L^{-1}})~ \mathrm{vs~EMF}:
y = 59.8x +600.8$$
  $0.7540~\mathrm{g}$ of toothpaste was diluted in a$50~\mathrm{mL}$ and tested to obtain a EMF value of $390$. what is the ppm of $\ce{F-}$ in $198~\mathrm{g}$ of toothpaste?   


I plugged in $390$ into the eqn to get $-3.525$, taking the antilog to obtain $0.0002985$, and times it by the volume $0.05~\mathrm{L}$ to get the moles of $\ce{NaF}$.

Due to stoichiometry, this also equals $1~\mathrm{mol}$ of $\ce{F-}$. I multiplied by $18.998~\mathrm{g~mol^{-1}}$ to obtain $0.000283527~\mathrm{mol}$ of $\ce{F-}$. I then multiplied by the factor: $\frac{198}{0.750}$ to obtain $0.074454015~\mathrm{g}$ of $\ce{F-}$ in the $198~\mathrm{g}$ of toothpaste.

I'm stuck here, how can I calculate the ppm?

The expected value of fluoride in toothpaste is $1000~\mathrm{ppm}$
Working with your numbers gives me the same result through moles of $\ce{NaF}$.  However, at this point you can multiply by the molar mass of fluoride, 18.998 g/mol, to get grams of fluoride, not moles, specifically, 0.0002835 g $\ce{NaF}$.

Parts per million is a mass-based concentration measure.  It's similar in concept to percent mass, but instead of parts per hundred, it uses parts per million.  Since we know the original mass of this sample, we can calculate the concentration in ppm based on that value.  It will be the same concentration as in the larger sample.

I get the following:
$\frac{0.0002835 \textrm{ g } \ce{F-}}{0.7540 \textrm{ g toothpaste}}\times 1000000 = 376.0\textrm{ ppm }\ce{F-}$