3 Exponential decrease
Consider the value, $\pounds D$ , of a car subject to depreciation, in terms of the age $A$ years of the car. The car was bought for $\pounds 10500$ . 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

$D$
had a fixed value of
$\pounds 10500$
when
$A$ = 0,
 $D$ decreases as $A$ increases and

$D$
decreases faster when
$A$
is small than when
$A$
is large. A plot of this function is shown in Figure 8.
Figure 8 :
Task!
Produce the linearised model of $D=10500{e}^{0.25A}$ .
$lnD=ln10500+ln\left({e}^{0.25A}\right)$
so $lnD=ln105000.25A$