A few problems

2

Silly things really, it's just gone from my head at the moment.

1.) The value of a second hand car reduces exponentially with age, so that its value $y after t years can be modelled by the formula

y=Ae^(-at)

If the car was $50k when new, and worth $38k after two years, find the values of A and a.

Use this to predict the value of the car when it is 5 years old.

2.) Solve:

a.) 10^(x-1)=3

3.) Using rules of indices simplify:

a.) x²y/xy^-1

b.) (xy^0.5)^4

Any step-by-step explanations would be great, cheers.

Reply 1

yusufu

16

1.) The value of a second hand car reduces exponentially with age, so that its value $y after t years can be modelled by the formula

y=Ae^(-at)

If the car was $50k when new, and worth $38k after two years, find the values of A and a.

t=0, y=A = 50000
t = 2,
38000 = 50000e^-2a
ln(19/25) = -2a
a = -½ ln(19/25)

Use this to predict the value of the car when it is 5 years old.

y = 50000e^-(-½t ln(19/25))
sub for t=5 etc.

2.) Solve:

a.) 10^(x-1)=3

log₁₀ 3 = x+1
x = log₁₀ 3 +1

3.) Using rules of indices simplify:

a.) x²y/xy^-1
xy²

b.) (xy^0.5)^4
x^4 y²

i accept rep :biggrin:

Reply 2

super_baros

Slice'N'Dice

Silly things really, it's just gone from my head at the moment.

1.) The value of a second hand car reduces exponentially with age, so that its value $y after t years can be modelled by the formula

y=Ae^(-at)

If the car was $50k when new, and worth $38k after two years, find the values of A and a.

Use this to predict the value of the car when it is 5 years old.

2.) Solve:

a.) 10^(x-1)=3

3.) Using rules of indices simplify:

a.) x²y/xy^-1

b.) (xy^0.5)^4

Any step-by-step explanations would be great, cheers.

1.The car is new.. so t=0
∴50000=A
after two years, t = 2
∴38000=Ae^(-at)
A=50000
39000/50000=e^-2a
19/25=e^-2a
25/19=e^2a
ln(25/19)=2a
a=ln(25/19)/2 = 0.137

Value of car when 5 years old...
t=5
y = 50000e^-5x0.137=$25204

2a)10^(x-1)=3
log10^(x-1)=log3
(x-1)log10=log3
x=log3/log10 + 1

3a)x²y/(xy^-1
x²y/xy^-1
xy/ 1/y
xy²

b.) (xy^0.5)^4
=x^4y²

I think they're right....