Answer to Question #209509 in Algebra for Nick

Question

it is observed that the number of bacteria present in the portion of the small intestine of an animal grows exponentially, initially there are 10,000 bacteria present. Three hours later, the bacterial growth to 500,000.Assuming no bacteria in the process, how many bacteria will be there be after one day? after how many hour will be bacteria double its number?

Answer 1

It follows that the equation for the number of bacteria present in the portion of the small intestine of an animal is $n(t)=Ca^t.$ Since initially for $t=0$ there are 10,000 bacteria, we have that $10,000=Ca^0=C,$ and hence $n(t)=10,000a^t$. Taking into account that for $t=3$ we have that $n(t)=500,000,$ we conclude that $500,000=10,000a^3,$ and thus $a=\sqrt[3]{50}$. We conclude that $n(t)=10,000(\sqrt[3]{50})^t=10,000\cdot50^{\frac{t}{3}}.$ Since $n(24)=10,000\cdot50^{\frac{24}{3}}=10,000\cdot50^{8}=5\cdot 10^{12},$ we conclude that after one day will be $5\cdot 10^{12}$ bacteria. If $n=2\cdot 10,000=20,000,$ then $20,000=10,000\cdot50^{\frac{t}{3}}.$ It follows that $50^{\frac{t}{3}}=2,$

and hence $t=3\cdot\log_{50}2\approx0.53\ hours\approx 32\ min.$ We conclude that arter 32 minutes bacteria double its number.