## College Algebra 7th Edition

$57$ minutes
The algebraic solution is to figure out how many amoeba can fit in the container. The amoeba splits into two every three minutes, so we can define the amount of amoeba in the tank as $f(x)=2^{x}$, where $x$ is the amount of three minute intervals. One hour has $20$-three minute periods, so we get $f(20)=2^{20}=1048576$ amoeba. Now, if two amoeba are placed, we have to change the function to show double the amount of amoeba. We set the function as $f(x)=2*2^{x}=2^{x+1}$ where $x$ is the amount of three minute intervals. We now solve for $x$ in the equation $f(x)=1048576=2^{x+1}$. We have said that $1048576=2^{20}$, substituting this into the equation yields $2^{20}=2^{x+1}$. This means that $20=x+1$, and $x=19$. Since there are $19$-three minute periods, there is a total of $19*3$, or $57$ minutes.