Question 331596
Initially, there are 5 bugs. If it quadruples in the first 9 minutes, there are 20 bugs.

{{{20=5*e^(k*9)}}}

{{{k=(2*ln(2))/9}}}

So, we have:



{{{5*e^((2*ln(2)*t)/9)=5*2^(2t/9)}}}

{{{700=5*2^(2t/9)}}}

{{{140=2^(2t/9)}}}

{{{t=32.08}}} minutes