## 2 Answers

N(t) &= N_0 \cdot (1 + r)^t \\

N(10) &= 10 \cdot N_0 \\

N(t) &= 2 \cdot N_0 \\

\frac{2 \cdot N_0}{10 \cdot N_0} &= \frac{(1 + r)^t}{(1 + r)^{10}} \\

\frac{1}{5} &= \left(\frac{1 + r}{(1 + r)^{10}}\right)^t \\

\left(\frac{1}{5}\right)^{\frac{1}{10}} &= 1 + r \\

r &= \left(\frac{1}{5}\right)^{\frac{1}{10}} - 1 \\

2 &= (1 + r)^t \\

\ln(2) &= \ln((1 + r)^t) \\

\ln(2) &= t \cdot \ln(1 + r) \\

t &= \frac{\ln(2)}{\ln(1 + r)} \\

t &\approx \frac{\ln(2)}{\ln\left(\left(\frac{1}{5}\right)^{\frac{1}{10}} - 1 + 1\right)} \\

t &\approx 3.32 \text{ hours}

\end{align*}

##
reply

N(t) = N0 * (1 + r)^t

Where N(t) is the number of bacteria after time t, N0 is the initial number of bacteria, r is the growth rate per unit of time, and t is the time.

We know that the number of bacteria increased tenfold in 10 hours, so:

N(10) = 10 * N0 N0 = initial number of bacteria

We need to find the time it takes for the number of bacteria to double, so:

N(t) = 2 * N0

Now, we can divide the second equation by the first equation to eliminate N0 and find the ratio of the growth rates:

(2 * N0) / (10 * N0) = [(1 + r)^t] / [(1 + r)^10]

Simplifying the equation, we get:

1/5 = (1 + r)^t / (1 + r)^10

Taking the 10th root of both sides to isolate the term with t:

(1/5)^(1/10) = 1 + r

Now, we can solve for r:

r = (1/5)^(1/10) - 1

Once we have the growth rate (r), we can plug it back into our equation for N(t) to find the time (t) when the number of bacteria doubles:

2 * N0 = N0 * (1 + r)^t

Divide both sides by N0:

2 = (1 + r)^t

Now take the logarithm of both sides (base doesn't matter as long as it's the same on both sides, so we'll use the natural logarithm, ln):

ln(2) = ln((1 + r)^t)

Using the logarithm property ln(a^b) = b * ln(a):

ln(2) = t * ln(1 + r)

Finally, solve for t:

t = ln(2) / ln(1 + r)

Plug in the value of r we found earlier:

t \(\approx\) ln(2) / ln((1/5)^(1/10) - 1 + 1)

t \(\approx\) 3.32 hours

So, it takes approximately 3.32 hours for the number of bacteria to double.