QUESTIONS - DIFFERENTIATION

QUESTIONS - ANSWERS - SHERILL - DIFFERENTIATION
sherill has left a new comment on your post "MATHS - 7.INDUCTION":
>
> Please Please Help! After T years, the value of a car purchased for $20,000
> is v=20,000(.75)^T When will the car be worth $500.00? Also, Use the
> derivative rule for y=b^x and find the derivative (rate of change) of the
> value of the car when: T=1 year and T=4 years

V = 20000*(0.75)^T=500
0.75^T = 500/20000=1/40
T*LOG 0.75 = -LOG(40)
T = -LOG(40)/LOG(0.75) = 12.823 YEARS.
DV/DT = 20000*(0.75^T)LOG(0.75)
WHEN T=1,WE GET
DV/DT = 20000*0.75*LOG(0.75)= -1874.1
WHEN T=4 , WE GET
DV/DT = 20000*(0.75^4)LOG(0.75)=-790.63