Answer
(a). $n(t)=25\times 2^{t/5}$
(b).$n(t)=303.14$
(c).$t=76.44$
Work Step by Step
$n(t)=n_0\times2^{t/a}$, whereas $n_0$: is the number of an Initial bacteria, $t$: time and $a$: is a time it takes to double.
(a).
Thus, In this case. $n_0=25$, $a=5 hr$.
$n(t)=25\times 2^{t/5}$
(b). $t=18$.
$n(t)=25\times 2^{18/5},$
$n(t)=303.14$
(c).$n(t)=1,000,000$,
$40000=2^{t/5},$
$\log (4\times10^4)=\frac{t}{5} \log 2$
$t=\frac{5\times(\log (4\times 10^4)}{\log 2} = 76.44$
