Answer:
![2000=1000 (8)^t](https://tex.z-dn.net/?f=%202000%3D1000%20%288%29%5Et)
If we dicide both sides by 1000 we got:
![2 = 8^t](https://tex.z-dn.net/?f=%202%20%3D%208%5Et)
Now we can apply natural log on both sides and we got:
![ln (2)= ln (8^t) = t ln(8)](https://tex.z-dn.net/?f=%20ln%20%282%29%3D%20ln%20%288%5Et%29%20%3D%20t%20ln%288%29)
And solving for t we got:
![t = \frac{ln(2)}{ln(8)}=0.333](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7Bln%282%29%7D%7Bln%288%29%7D%3D0.333)
So then we can conclude that every 0.333 seconds the amount of bacteria is doubled under the model assumed.
The number of bacteria is doubled every 0.333 seconds
Step-by-step explanation:
For this case we have the following model given for the number of bacteria after t seconds:
![N(t) = 1000 (8)^t](https://tex.z-dn.net/?f=%20N%28t%29%20%3D%201000%20%288%29%5Et)
We want to find the time that is required to double the number of bacteria. If we analyze the function we see that the initial amount for t=0 is 1000. And if we want to double this amount we need to have 2000. And using this we have:
![2000=1000 (8)^t](https://tex.z-dn.net/?f=%202000%3D1000%20%288%29%5Et)
If we dicide both sides by 1000 we got:
![2 = 8^t](https://tex.z-dn.net/?f=%202%20%3D%208%5Et)
Now we can apply natural log on both sides and we got:
![ln (2)= ln (8^t) = t ln(8)](https://tex.z-dn.net/?f=%20ln%20%282%29%3D%20ln%20%288%5Et%29%20%3D%20t%20ln%288%29)
And solving for t we got:
![t = \frac{ln(2)}{ln(8)}=0.333](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7Bln%282%29%7D%7Bln%288%29%7D%3D0.333)
So then we can conclude that every 0.333 seconds the amount of bacteria is doubled under the model assumed.
The number of bacteria is doubled every 0.333 seconds