Question 1169051
Let the initial production of the lathe be $P_0 = 10000$.

**a. Production in one hour:**

One hour has 60 minutes. The machine doubles its production every 20 minutes. So, in one hour, the production will double $60 / 20 = 3$ times.

* After the first 20 minutes: $P_1 = P_0 \times 2 = 10000 \times 2 = 20000$
* After the next 20 minutes (total 40 minutes): $P_2 = P_1 \times 2 = 20000 \times 2 = 40000$
* After the final 20 minutes (total 60 minutes or one hour): $P_3 = P_2 \times 2 = 40000 \times 2 = 80000$

Alternatively, we can use the formula:
$P(t) = P_0 \times 2^{t/d}$
where:
* $P(t)$ is the production after time $t$
* $P_0$ is the initial production
* $t$ is the total time
* $d$ is the doubling time

For one hour (60 minutes):
$P(60) = 10000 \times 2^{60/20}$
$P(60) = 10000 \times 2^3$
$P(60) = 10000 \times 8$
$P(60) = 80000$

So, the machine will produce **80,000** units in one hour.

**b. Time to produce 8 million:**

We want to find the time $t$ when the production $P(t)$ reaches 8,000,000.
$P(t) = 10000 \times 2^{t/20} = 8000000$

Divide both sides by 10000:
$2^{t/20} = \frac{8000000}{10000}$
$2^{t/20} = 800$

To solve for $t$, we can take the logarithm of both sides (using base 2 or natural logarithm):

Using base 2 logarithm:
$\log_2(2^{t/20}) = \log_2(800)$
$\frac{t}{20} = \log_2(800)$

We know that $2^9 = 512$ and $2^{10} = 1024$. So, $\log_2(800)$ is between 9 and 10.
$\log_2(800) = \log_2(8 \times 100) = \log_2(2^3 \times 100) = 3 + \log_2(100)$
Since $2^6 = 64$ and $2^7 = 128$, $\log_2(100)$ is between 6 and 7 (approximately 6.64).
$\log_2(800) \approx 3 + 6.64 = 9.64$

Now, solve for $t$:
$t = 20 \times \log_2(800)$
$t \approx 20 \times 9.64$
$t \approx 192.8$ minutes

Alternatively, using natural logarithm:
$\ln(2^{t/20}) = \ln(800)$
$\frac{t}{20} \ln(2) = \ln(800)$
$t = 20 \times \frac{\ln(800)}{\ln(2)}$
$t \approx 20 \times \frac{6.6846}{0.6931}$
$t \approx 20 \times 9.644$
$t \approx 192.88$ minutes

So, it will take approximately **192.88 minutes** for the machine to produce 8 million units.

To express this in hours and minutes:
$192.88 \text{ minutes} = 3 \text{ hours and } 12.88 \text{ minutes}$
$0.88 \text{ minutes} \times 60 \text{ seconds/minute} \approx 53 \text{ seconds}$

So, it will take approximately 3 hours, 12 minutes, and 53 seconds.

Final Answers:
a. The machine will produce **80,000** units in one hour.
b. It will take approximately **192.88 minutes** (or about 3 hours, 12 minutes, and 53 seconds) for the machine to produce 8 million units.