Question

Jeremy bought a new truck for $32,000. The value of the truck after t years can be represented by the formula V = 32,000(.8)t. When will the truck be worth approximately $6700?

Answer 1

I'm assuming you mean for the equation to be
\[V=32,000(.8)^t\]
Because it doesn't work if it is a linear equation.
So if V represents the value of the truck then with our equation
\[V=32,0000(.8)^t\]
We plug in 6,700 for V and solve for t, because V is the future value of the truck, and t is the number of years after purchase of the truck that it will have the value of $6,700. So,
\[6700=32,000(.8)^t\]
We simplify by dividing both sides by 32,000 to get:
\[0.18108=(0.8)^t\]
From here we take the common, base 10 logarithm of both sides (though any other base works, as long as you log both sides with the same base).
\[log(0.18108)=t log(0.8)\]
The t moves from a power to coefficient of log(0.8) because a logarithm "undoes" a exponential function (this is also a property of logs).
From there, we divide both sides by log(0.8) to get:
\[\frac {\log(0.18108)}{\log(0.8)}=t\]
So then,
\[t=7.66\]
But years can only be whole numbers, so we round up to
\[t=8\]

Answer 2

Would this be an example of exponential decay?

Answer 3

Yes, because as time goes on, the value of something decreases, or decays.

Answer 4

Thanks!

Answer 5

No problem :)