Question
If a bacteria population doubles every 5 days, what is the growth rate compounded daily? Round your answer to one
decimal place.

Question

Question If a bacteria population doubles every 5 days, what is the growth rate compounded daily? Round your answer to one decimal place.

UpStudy AI · Accepted Answer

1. We know that the bacteria population doubles every 5 days. This can be written as:
   $$
   P(5) = 2P(0)
   $$

2. With a daily compounded growth rate $$ r $$, the population after 5 days is given by:
   $$
   P(5) = P(0)(1+r)^5
   $$

3. Equate the two expressions for $$ P(5) $$:
   $$
   P(0)(1+r)^5 = 2P(0)
   $$
   Dividing both sides by $$ P(0) $$ (assuming $$ P(0) \neq 0 $$):
   $$
   (1+r)^5 = 2
   $$

4. Taking the fifth root of both sides gives:
   $$
   1+r = 2^{\frac{1}{5}}
   $$

5. Solving for $$ r $$, we obtain:
   $$
   r = 2^{\frac{1}{5}} - 1
   $$

6. To approximate the value, we compute:
   $$
   2^{\frac{1}{5}} = \exp\left(\frac{\ln{2}}{5}\right) \approx \exp(0.1386) \approx 1.1487
   $$
   Thus,
   $$
   r \approx 1.1487 - 1 = 0.1487
   $$

7. Expressed as a percentage, the daily compounded growth rate is approximately:
   $$
   14.9\%
   $$

So, the growth rate compounded daily is $$ 14.9\% $$.
The daily compounded growth rate is approximately 14.9%.

Question
If a bacteria population doubles every 5 days, what is the growth rate compounded daily? Round your answer to one
decimal place.

Upstudy AI Solution

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