Question 190620
{{{b=100(2^t)}}} Start with the given equation.



{{{51200=100(2^t)}}} Plug in {{{b=51200}}} (the given bacteria population)



{{{512=2^t}}} Divide both sides by 100.



{{{log(10,(512))=log(10,(2^t))}}} Take the log base 10 of both sides



{{{log(10,(512))=t*log(10,(2))}}} Rewrite right side using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{log(10,(512))/log(10,(2))=t}}} Divide both sides by {{{log(10,(2))}}}



{{{log(2,(512))=t}}} Use the change of base formula to rewrite the left side



Note: Remember the change of base formula is {{{log(b,(x))=log(10,(x))/log(10,(b))}}}



{{{t=log(2,(512))}}} Rearrange the equation



{{{t=9}}} Evaluate the log base 2 of 512 to get 9



Note: 



since {{{2^9=512}}}, this means that {{{log(2,(512))=9}}}



So the answer is {{{t=9}}} which means that it will take 9 minutes for the population to reach 51,200