Revision as of 00:03, 26 November 2023 by Admin (Created page with "We want <math>\left(100(1+i)^2-50\right)(1+i)^n=200</math>. Thus <math>(1+i)^n=\frac{200}{\left(100(1+i)^2-50\right)}</math> so <math display="block"> n=\frac{\ln (200)-\ln \left(100(1+i)^2-50\right)}{\ln (1+i)} \text {. } </math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwindsor.ca | title = University of Windsor Old Tests 62-392 Theory of Interest | access-date=November 23, 2023}}")
Exercise
ABy Admin
Nov 26'23
Answer
We want [math]\left(100(1+i)^2-50\right)(1+i)^n=200[/math]. Thus [math](1+i)^n=\frac{200}{\left(100(1+i)^2-50\right)}[/math] so
[[math]]
n=\frac{\ln (200)-\ln \left(100(1+i)^2-50\right)}{\ln (1+i)} \text {. }
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.