Is my working correct in regards to this question? I'm quite stuck on it and I'm not too sure if I am in the right direction. Any advice is appreciated. Thank you.
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 in temperature from ambient decays exponentially.
Let the ambient temperature be $t°C$, then:
$(18-t) = (22-t)*b^5$, or $b^{5} = [(18-t)/(22-t)]^3$ ...... [ I ]
$(15-t)= (22-t)*b^{15}$, or $b^{15} = [(15-t)/(22-t)] $.... [ II ]
equating the two
$(15-t)(22-t)^2 = (18-t)^3$
opening out,
$5t ^2 -172t +1428 = 0$
factorising,
$(5x-102)(x-14) = 0$
taking value less than $15$,
$t = 14°C$.
Your solution is very correct and your reasoning is good.
Temperature varies exponentially as $$T_{read}(t)=T_{out}+(T_{in}-T_{out}) e^{-kt}$$ (you could write the exponential in another form, just as you did).
For $t=0$ we have $$T_{read}(0)=T_{in}=22$$ which does not bring much information. Now,$$T_{read}(5)=T_{out}+(T_{in}-T_{out}) e^{-5k}=18$$ $$T_{read}(15)=T_{out}+(T_{in}-T_{out}) e^{-15k}=15$$ So we have two equations for two unknowns $T_{out}$ and $k$ so you could know $T_{read}$ for any value of $t$.
I am sure that you can take from here using the trick you already used (that it to say that one of the exponential is the cube of the other).