Is my working correct? Exponenial decay

Question

Is my working correct? If not, please let me know where I have gone wrong. Thank you for taking the time to check!

Question: 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?

My working: the difference from ambient temperature $D$ decays exponentially.

let the outdoor temp be $t$, then in the form $D = ab^x$,

since it decays from $(22-t)$ to $(18-t)$ in 5 minutes,

$D(t) = (22-t)\times (\frac{18-t}{22-t})^{\frac{x}{5}}$

thus $(15-t) = (22-t)\times (\frac{18-t}{22-t})^{\frac{15}{5}}$

$(18-t) = (15-t)\times(22-t)^2$ this yields (within the feasible range) $t = 14.939$

Answer 1

The equation $$15-t=(22-t)\left(\frac{18-t}{22-t}\right)^3$$ is correct. (The equation $(18-t) = (15-t)\times(22-t)^2$ is not.)

The relevant root is $t=14$.

Answer 2

From your equation, $$ \ 15-t \ \ = \ \ (22-t) · \left(\frac{18-t}{22-t}\right)^3 \ \ \Rightarrow \ \ (18-t)^3 \ = \ (15-t) · (22-t)^2 \ \ , $$

it is possible to "eyeball" the solution for the ambient temperature, $ \ t = 14 \ \ . $ The implied quadratic equation, $ \ 5t^2 - 172t + 1428 \ = \ 0 \ \ $ is manageable with the aid of a calculator (the second solution, $ \ t = 20.4 \ $ is plainly spurious), but a less convenient set of conditions would make this strictly a job for machine aid.

Another approach suggested by your work would be to define a ratio $ \ r \ = \ \frac{18-t}{22-t} \ \Rightarrow \ \frac{15-t}{22-t} \ = \ r^3 \ \ . $ If we write $ \ \Delta \ = \ 22 - t \ \ , $ the information from the problem statement can be expressed as

$$ \mathbf{initial \ condition:} \quad t \ + \ \Delta \ = \ 22 \ \ ; \ \ \mathbf{at \ 5 \ minutes:} \quad t \ + \ r·\Delta \ = \ 18 \ \ ; $$ $$ \mathbf{at \ 15 \ minutes:} \quad t \ + \ r^3·\Delta \ = \ 15 \ \ . $$

Subtracting each of the latter two equations from the first one produces $$ (1-r)·\Delta \ = \ 4 \ \ \ , \ \ \ (1-r^3)·\Delta \ = \ 7 \ \ . $$

Dividing the second of these results by the first leads to

$$ \frac{1 \ - \ r^3}{1 \ - \ r} \ = \ \frac74 \ \ \Rightarrow \ \ 1 \ + \ r \ + r^2 \ \ = \ \frac74 \ \ \Rightarrow \ \ r^2 \ + \ r \ - \ \frac34 \ = \ 0 \ \ \Rightarrow \ \ r \ = \ \frac12 \ , \ -\frac32 \ \ , $$ the negative ratio being inapplicable. We may immediately conclude that $ \ \Delta \ = \ 8 \ \ $ and $ \ t \ = \ 14 \ . $

Either of these methods works this tidily only because the ratio of the time intervals from the "initial moment" at which the temperature measurements are made is $ \ 3:1 \ . \ $ A larger integer ratio or a non-integer ratio would tend to require a solution by numerical approximation.