Question Number 144049 by physicstutes last updated on 20/Jun/21
$$\:\mathrm{Given}\:\mathrm{the}\:\mathrm{equation}\:\:\mathrm{1000}\:=\:\mathrm{2000}\left(\frac{\mathrm{1}−\left(\mathrm{1}+{t}\right)^{−{n}} }{{t}}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{t}. \\ $$
Answered by Olaf_Thorendsen last updated on 21/Jun/21
$$\mathrm{1000}\:=\:\mathrm{2000}\left(\frac{\mathrm{1}−\left(\mathrm{1}+{t}\right)^{−{n}} }{{t}}\right) \\ $$$$\frac{\mathrm{1}−\left(\mathrm{1}+{t}\right)^{−{n}} }{{t}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{t}}.\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)^{{n}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{t}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${q}\left(\mathrm{1}−{q}^{{n}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{q}\right) \\ $$$$\mathrm{Let}\:{q}\:=\:\frac{\mathrm{1}}{\mathrm{1}+{t}} \\ $$$${q}^{{n}+\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{2}}{q}+\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{0} \\ $$