Thread: Calculating depreciation using given model?

1. Calculating depreciation using given model?

Depreciation model for the price (P) in of the car a person wishes to buy is given by P=45520e^-0.1t, where t is the number of years since the car was brand new. The savings model for the amount (A) in dollars, this person has accumulated in the same t years is given by A=10,000(1.05)^t.

How many years (to 2 dps) will it take for this person to be able to buy the car (ie. where A=P) using his savings only?

The answer in the book is 10.19 years, but i dont know how to get there.

2. $P = 45520e^{-0.1t}$

$A = 10000(1.05)^t$

When P = A;

$45520e^{-0.1t} = 10000(1.05)^t$

Can you solve this?

3. $45520e^{-0.1t} = 10000(1.05)^t \setminus \cdot \frac{e^{0.1t}}{10000}$
$(1.05)^t\cdot e^{0.1t}=\frac{45520}{10000}$
$(1.05\cdot e^{0.1})^t=4.552$
$\log \left((1.05\cdot e^{0.1})^t\right)=\log 4.552$
$t\cdot \log \left(1.05\cdot e^{0.1}\right)=\log 4.552$
$t=\frac{\log 4.552}{\log \left(1.05\cdot e^{0.1}\right)}\approx 10.1859$

4. Thank you MathoMan, however can you please explain the first line (in particular \.e^.1t/10000)

5. This is how I do it:

$45520e^{-0.1t} = 10000(1.05)^t$

$ln(45520e^{-0.1t}) = ln(10000(1.05)^t)$

$ln(45520) - 0.01tln(e) = ln(10000) + tln(1.05)$

$ln(45520) - ln (10000) = tln(1.05) + 0.01t$

$ln(45520) - ln (10000) = t(ln(1.05) + 0.01)$

$t = \frac{ln(45520) - ln (10000)}{ln(1.05) + 0.01}$

$t = \frac{ln(4.5520)}{ln(1.05) + 0.01}$

$t = 10.185$

6. Thank you heaps

7. Originally Posted by daniel52947
Thank you MathoMan, however can you please explain the first line (in particular \.e^.1t/10000)
It means that the whole equation is multiplied by the expression $\frac{e^{0.1t}}{10000}$