### 3 Exponential decrease

Consider the value, $£D$ , of a car subject to depreciation, in terms of the age $A$ years of the car. The car was bought for $£10500$ . The function

$\phantom{\rule{2em}{0ex}}D=10500{e}^{-0.25A}\phantom{\rule{2em}{0ex}}\left(0\le A\le 6\right)$

could be considered appropriate on the ground that

1. $D$ had a fixed value of $£10500$ when

$A$ = 0,

2. $D$ decreases as $A$ increases and
3. $D$ decreases faster when $A$ is small than when $A$ is large. A plot of this function is shown in Figure 8.

Figure 8 :

Produce the linearised model of $D=10500{e}^{-0.25A}$ .

$lnD=ln10500+ln\left({e}^{-0.25A}\right)$

so $lnD=ln10500-0.25A$