Does anyone know how I would go about answering this question? Any feedback is appreciated!
A thermometer that has been stored indoors where the temperature is 22 degrees Celsius, is taken outdoors. After 5 minutes it reads 18 degrees. After 15 minutes it reads 15 degrees. What is the outdoor temperature?
The temperature reading - $T$ - will increase or decrease exponentially with time - $t$ - from the inside temperature - $T_i$ - towards the true temperature - $T_o$. This means
$$T(t) = T_o - (T_o-T_i) \exp(-\lambda t)$$
We can use the given values to solve for $T_o$. We know already that $T_i=22$ so there are two unknowns ($T_o$ and $\lambda$) but we have two values so we have two equations.
$$T(5)=T_o-(T_o-T_i)\exp(-\lambda 5)$$
which rearranges to give
$$\lambda=\frac{1}{-5}\ln\left(\frac{T_o-T(5)}{T_o-Ti}\right)$$
Exactly the same can be done for $T(15)$ which gives
$$\lambda=\frac{1}{-15}\ln\left(\frac{T_o-T(15)}{T_o-Ti}\right)$$
Equating these two and solving for $T_o$
$$T_o = \frac{ T(15)\,e^{\frac{1}{3}}-T(5)}{e^{\frac{1}{3}}-1}$$
Plugging in the values gives us therefore that the outdoor temperature is $T_o=7.42$ ($3$ s.f.).