Answer :
We are given the oven cooling function
$$
f(t) = 349.2 \times (0.98)^t,
$$
where $t$ is the time in minutes that the oven has been cooling. We want to determine:
1. The temperature of the oven after 15 minutes.
2. The time needed for the temperature to drop to half of its initial value.
Let's go through each part step-by-step.
----------------------------------------
**1. Temperature after 15 minutes**
The temperature after $t = 15$ minutes is given by
$$
f(15) = 349.2 \times (0.98)^{15}.
$$
Evaluating this expression yields
$$
f(15) \approx 257.9083.
$$
Thus, after 15 minutes, the oven's temperature is approximately $257.9083$ (in the same units as the initial temperature).
----------------------------------------
**2. Time for the temperature to drop to half the initial value**
The initial temperature of the oven is $349.2$. Half of this value is
$$
\frac{349.2}{2} = 174.6.
$$
We need to find the time $t$ when
$$
349.2 \times (0.98)^t = 174.6.
$$
To solve for $t$, first divide both sides of the equation by $349.2$:
$$
(0.98)^t = \frac{174.6}{349.2} = 0.5.
$$
Now, take the natural logarithm of both sides:
$$
\ln\left((0.98)^t\right) = \ln(0.5).
$$
Using the logarithm power rule, $\ln(a^b) = b \ln a$, we obtain:
$$
t \ln(0.98) = \ln(0.5).
$$
Solving for $t$ gives:
$$
t = \frac{\ln(0.5)}{\ln(0.98)}.
$$
Evaluating this expression yields
$$
t \approx 34.3096.
$$
Thus, it takes approximately $34.3096$ minutes for the oven's temperature to drop to half of its initial value.
----------------------------------------
**Summary**
- The temperature after 15 minutes is approximately $257.9083$.
- The time required for the temperature to reach half of its initial value is approximately $34.3096$ minutes.
$$
f(t) = 349.2 \times (0.98)^t,
$$
where $t$ is the time in minutes that the oven has been cooling. We want to determine:
1. The temperature of the oven after 15 minutes.
2. The time needed for the temperature to drop to half of its initial value.
Let's go through each part step-by-step.
----------------------------------------
**1. Temperature after 15 minutes**
The temperature after $t = 15$ minutes is given by
$$
f(15) = 349.2 \times (0.98)^{15}.
$$
Evaluating this expression yields
$$
f(15) \approx 257.9083.
$$
Thus, after 15 minutes, the oven's temperature is approximately $257.9083$ (in the same units as the initial temperature).
----------------------------------------
**2. Time for the temperature to drop to half the initial value**
The initial temperature of the oven is $349.2$. Half of this value is
$$
\frac{349.2}{2} = 174.6.
$$
We need to find the time $t$ when
$$
349.2 \times (0.98)^t = 174.6.
$$
To solve for $t$, first divide both sides of the equation by $349.2$:
$$
(0.98)^t = \frac{174.6}{349.2} = 0.5.
$$
Now, take the natural logarithm of both sides:
$$
\ln\left((0.98)^t\right) = \ln(0.5).
$$
Using the logarithm power rule, $\ln(a^b) = b \ln a$, we obtain:
$$
t \ln(0.98) = \ln(0.5).
$$
Solving for $t$ gives:
$$
t = \frac{\ln(0.5)}{\ln(0.98)}.
$$
Evaluating this expression yields
$$
t \approx 34.3096.
$$
Thus, it takes approximately $34.3096$ minutes for the oven's temperature to drop to half of its initial value.
----------------------------------------
**Summary**
- The temperature after 15 minutes is approximately $257.9083$.
- The time required for the temperature to reach half of its initial value is approximately $34.3096$ minutes.
