Question

A typical car depreciates in value by about 15% per year. How long would it take a $35000 car to be valued at $15000

Answer 1

solve
\[15,000=35000(.85)^t\] for t

Answer 2

so subtract the 15000 from 35000?
or?

Answer 3

\[\frac{3}{7}=.85^t\]
\[t=\frac{\ln(\frac{3}{7})}{\ln(.85)}\]

Answer 4

first divide, then use change of base

Answer 5

divide by 3500?

Answer 6

yes divide by 35000

Answer 7

wait, whats the formula you used?

Answer 8

that is where i got the
\[\frac{3}{7}\] from

Answer 9

lets start at the beginning.
the car loses 15% of its value every year, so every year it is worth 85% of the previous years value. that is how i got
\[35,000\times (.85)^t\]

Answer 10

now you want to know when it is worth 15,000 so put
\[15,000=35,000\times (.85)^t\] and solve for t
first step is to divide by 35,000 to get
\[\frac{3}{7}=.85^t\] so far so good?

Answer 11

yes. the whole lm thing confuses me

Answer 12

just like in the next question you posted, so solve
\[b^t=A\] for t use
\[t=\frac{\ln(A)}{\ln(b)}\]

Answer 13

so if i see
\[.85^t=\frac{3}{7}\] i solve for t in on step and write
\[t=\frac{\ln(\frac{3}{7})}{\ln(.85)}\]

Answer 14

thank you sooo much. so now how do i solve the other problem that i posted.