Hi,

I've been asked the following question:

The temperature of an electrical device at a time 't' is given by the function:

$\displaystyle {T}_{2} = {T}_{1}(1-{e}^{\frac{-t}{\tau}}) $

Where $\displaystyle {T}_{1} $ is the temperature at time t = 0 and $\displaystyle \tau $ is a constant

Find a) $\displaystyle {T}_{1} $ when $\displaystyle {T}_{2} $ is 60 degrees Celsius, t = 50 seconds and $\displaystyle \tau $ = 90 seconds

b) The time t, for $\displaystyle {T}_{2} $ to be $\displaystyle \frac{1}{3} $ the value of $\displaystyle {T}_{1} $

I believe the answer for (a) to be:

$\displaystyle \frac{60}{1-{e}^{\frac{-50}{90}}} = {T}_{1} = 140.763593 $

However for question (b) I've gotten:

$\displaystyle 46.92119767 = 140.763593(1-{e}^{\frac{-t}{90}})$

$\displaystyle \frac{46.92119767}{140.763593} = 1-{e}^{\frac{-t}{90}}$

$\displaystyle (-1)\frac{46.92119767}{140.763593} = (-1)1-{e}^{\frac{-t}{90}}$

$\displaystyle -\frac{46.92119767}{140.763593} = -1+{e}^{\frac{-t}{90}}$

$\displaystyle -\frac{46.92119767}{140.763593} + 1 = -1+1+{e}^{\frac{-t}{90}}$

$\displaystyle -\frac{46.92119767}{140.763593} + 1 = {e}^{\frac{-t}{90}}$

$\displaystyle ln(-\frac{46.92119767}{140.763593} + 1) = ln({e}^{\frac{-t}{90}})$

$\displaystyle ln(-\frac{46.92119767}{140.763593} + 1) = \frac{-t}{90}$

$\displaystyle ln(-\frac{46.92119767}{140.763593} + 1)(90) = -t$

$\displaystyle (-1)ln(-\frac{46.92119767}{140.763593} + 1)(90) = (-1)-t$

$\displaystyle (-1)ln(-\frac{46.92119767}{140.763593} + 1)(90) = t$

$\displaystyle t = 36.49185973 $ Seconds

Can someone please let me know if I've gotten these correct?

Thanks.