How Can Exponential Decay Reveal Initial Temperature?

[Boba Fet]

Suppose that you take a thermometer outside where it is 100°.

T(5min)=80° T(15min)=90°

What is the initial temp of the thermometer?

Given equation

dT/d t= k(T-Te)

Derived Equation
⌠(T-Te)^-1 (dT/dt)dt =⌠ kdt
ln(T-Te)=kt + c
T=ce^kt + Te

so i basically got the answer by knowing c must be negative and that when i use a square root on the magnitude of c i through out the positive value.
Wolfram Alpha
my real question is how can this become -3 ln(-20/x) +ln(-10/x) becomes -2ln(-(1/x)) -ln(800)?

Attempt to find C

1) 80=ce^k5 +100 ---- -20=ce^k5 ----- -20/c=e^k5 ------ ln(-20/c)=k5 ----- -3ln(-20/c)=-k15

2) 90=ce^k15 +100 ---- -10=ce^k15 ----- -10/c=e^k15 ------ ln(-10/c)=k15

becomes

ln(-10/c) - 3ln(-20/c) ----- ln((-10/c) x (c^3 / -8000)) -------- ln(c^2/800) ---- 2ln(c) -ln(800)=0

ln(c)=(1/2)ln(800)

c= +/- sqrt(800)

 

[Simon Bridge]

my real question is how can this[?] become -3 ln(-20/x) +ln(-10/x) becomes -2ln(-(1/x)) -ln(800)?

Are you saying that the expression:
$-3\ln\frac{-20}{x} + \ln \frac{-10}{x} = -2\ln\frac{-1}{x} - \ln 800$ ... is not true?
... or you don't know how to get there from

T=ce^kt + Te

... which I am reading as $T(t)=Ce^{kt} + T_e$?

The rest of your post appears to show you doing the calculation (though you need to say what finding "C" does for you.)

 

[Boba Fet]

yes i don't know how to get there

− 3 ln − 20 x + ln − 10 x = − 2 ln − 1 x − ln 800 −3ln⁡−20x+ln⁡−10x=−2ln⁡−1x−ln⁡800

finding c will get me the initial temp bc t will cancel out k when t is 0.Reference https://www.physicsforums.com/threads/find-this-initial-temperature.887268/

 

[Simon Bridge]

Do you know the relationship between the initial temperature and C?

yes i don't know how to get there
− 3 ln − 20 x + ln − 10 x = − 2 ln − 1 x − ln 800 −3ln⁡−20x+ln⁡−10x=−2ln⁡−1x−ln⁡800

... that last bit does not make sense.
Off post #1, the way to understand how to get "there" must start from knowing what "x" stands for.
However, if you can find the initial temperature without going "there", then why bother?

 

[Chestermiller]

Your solution to the differential equation should read:$$(T_e-T)=(T_e-T_0)e^{-kt}$$
So, $$20=(T_e-T_0)e^{-5k}\tag{1}$$
$$10=(T_e-T_0)e^{-15k}\tag{2}$$If you cube Eqn. 1, you get:$$8000=(T_e-T_0)^3e^{-15k}\tag{3}$$
What do you get if you divide Eqn. 3 by Eqn. 2?