T(t) = Ta + (To - Ta)e^kt

T(10) = 35 = 70 + (28-70)e^10k

e^10k = ln(-35/-42) = ln(5/6)

k = (1/10)*ln(5/6) = -.018232

T(30) = 70 + (28-70)e^(-.018232*30) = 45.7° <-------

45 = 70 + (28-70)e^(-.018232t)

-.018232t = 25/42

ln(25/42)/-.018232 = 28.46 min <---------

NOTE: If you are not compelled to use "e" and k", you can do much more simply with less chances of error, using the crux of the law, viz. the DIFFERENCE (D) from ambient decays exponentially //

Every 10 mins, D decays to 5/6 of previous value ,

so D(t) = 42*(5/6)^(0.1t) ,

D(30) = 42*(5/6)^3 = 24.3,

T(30) = 70 - 24.3 = 45.7° <------

25 = 42*(5/6)^(0.1t)

t = 10*ln(25/42)/ln(5/6) = 28.46min <------

Newton's law of cooling ...dN / dt = k [ 70 - N ] , N(0) = 28 ; N(10) = 35...

N = C e^(kt)+ 70......C = - 42.......N(10) = 35 = -42 e^(10k) + 70----> e^k = [5/6]^(1/10)

N(t) = - 42 [ 5/6]^(0.1 t) + 70......let t = 30 , then let N = 45 and find t