Question Number 205915 by dodo last updated on 02/Apr/24
![](https://www.tinkutara.com/question/33630.png)
Answered by Skabetix last updated on 03/Apr/24
![Placement : U_n = 6 000 000 ×1.03^n Valeur vehicule : V_n = 30 000 000 ×0.95^n En notant n le nombre d annees on a : 6 000 000 × 1.03^n =30 000 000 × 0.95^n ⇔(((1.03)/(0.95)))^n = 5 ⇔nln(((1.03)/(0.95)))=ln(5) ⇔n=((ln(5))/(ln(((1.03)/(0.95))))) ⇒ n =20 Donc au bout de 20 ans](https://www.tinkutara.com/question/Q205936.png)
$${Placement}\::\:{U}_{{n}} =\:\mathrm{6}\:\mathrm{000}\:\mathrm{000}\:×\mathrm{1}.\mathrm{03}^{{n}} \\ $$$${Valeur}\:{vehicule}\::\:{V}_{{n}} =\:\mathrm{30}\:\mathrm{000}\:\mathrm{000}\:×\mathrm{0}.\mathrm{95}^{{n}} \\ $$$${En}\:{notant}\:{n}\:{le}\:{nombre}\:{d}\:{annees}\:{on}\:{a}\:: \\ $$$$\mathrm{6}\:\mathrm{000}\:\mathrm{000}\:×\:\mathrm{1}.\mathrm{03}^{{n}} =\mathrm{30}\:\mathrm{000}\:\mathrm{000}\:×\:\mathrm{0}.\mathrm{95}^{{n}} \\ $$$$\Leftrightarrow\left(\frac{\mathrm{1}.\mathrm{03}}{\mathrm{0}.\mathrm{95}}\right)^{{n}} =\:\mathrm{5} \\ $$$$\Leftrightarrow{nln}\left(\frac{\mathrm{1}.\mathrm{03}}{\mathrm{0}.\mathrm{95}}\right)={ln}\left(\mathrm{5}\right) \\ $$$$\Leftrightarrow{n}=\frac{{ln}\left(\mathrm{5}\right)}{{ln}\left(\frac{\mathrm{1}.\mathrm{03}}{\mathrm{0}.\mathrm{95}}\right)} \\ $$$$\Rightarrow\:{n}\:=\mathrm{20} \\ $$$${Donc}\:{au}\:{bout}\:{de}\:\mathrm{20}\:{ans} \\ $$