Question 726652
{{{B=94000*5^x}}}
When {{{x=0}}} --> {{{B=94000*5^0=94000*1=94000}}}
When {{{x=4}}} --> {{{B=94000*5^4=94000*625=5875000}}}
 
EXTRA:
If I did not have a calculator, I would still know that C) is the right choice, because
When {{{x=4}}} --> {{{B=94000*5^4=(47*100*2)*(5*5^3)=47*(100*2*5)*5^3=47*1000*5^3=47*5^2*5000}}}
and 5,875,00 is the only multiple of 5,000 in the lot.