Question

Depreciation The value V of a Dodge Stratus that is t years old can be modeled by V(t) = 16,775(0.905)t.
(a) According to the model, when will the car be worth $15,000?
(b) According to the model, when will the car be worth $4000?

Answer 1

\[\Large\rm V(t)=16775(0.905)^t\]Was the t supposed to be an exponent like this?

Answer 2

So according to this model, the car is depreciating to 90.5% of it's previous value each year.

For part (a):
V is for value. They want to know when the value, V, will be equal to 15000.\[\Large\rm 15000=16775(0.95)^t\]Then we have to solve for t.
If you haven't done a lot of work with exponential functions then this can be a little tricky.

Answer 3

We'll first isolate the exponential portion by dividing the 16775 to the other side,\[\Large\rm \frac{15000}{16775}=(0.95)^t\]We'll take the natural log of each side,
err before we do that, let's simplify the fraction on the left,\[\Large\rm 0.894=(0.95)^t\]Taking the natural log of each side,\[\Large\rm \ln 0.894=\ln (0.95)^t\]

Answer 4

The reason we do this is because there is a log rule that allows us to take thing out of the exponent position.

\[\Large\rm \ln(a^b)=b \ln(a)\]

Answer 5

So we can bring the t down,\[\Large\rm \ln 0.894=t~\ln (0.95)\]To finish solving for t, we'll divide this ugly log to the other side,\[\Large\rm t=\frac{\ln(0.894)}{\ln(0.95)}\]

Answer 6

Just be careful plugging this into your calculator.
Use brackets whenever you're not sure.
So you would plug it in something like this:\[\Large\rm \ln(0.894)\div \ln(0.95)\]Ehh I guess this one is pretty straight forward actually.

Answer 7

Thank you I did not use brackets ;)