Solve the problem.
-The number of bacteria growing in an incubation culture increases with time according to $\mathrm { B } ( \mathrm { x } ) = 9800 ( 2 ) ^ { \mathrm { x } }$, where $x$ is time in days. Find the number of bacteria when $x = 0$ and $x = 3$.
A) 19,$600 ; 78,400$
B) $9800 ; 58,800$
C) $9800 ; 39,200$
D) $9800 ; 78,400$

Question

Quizplus · Accepted Answer

To find the number of bacteria when $x=0$, we need to substitute $x=0$ into the given equation:

$$\mathrm { B } (0) = 9800(2)^0 = 9800$$

So there are 9,800 bacteria when $x=0$.

To find the number of bacteria when $x=3$, we need to substitute $x=3$ into the given equation:

$$\mathrm { B } (3) = 9800(2)^3 = 78,400$$

So there are 78,400 bacteria when $x=3$.

Therefore, the answer is D.

Solve the Problem B(x)=9800(2)x\mathrm { B } ( \mathrm { x } ) = 9800 ( 2 ) ^ { \mathrm { x } }B(x)=9800(2)x

Solve the Problem $\mathrm { B } ( \mathrm { x } ) = 9800 ( 2 ) ^ { \mathrm { x } }$