Revision as of 17:25, 26 November 2023 by Admin (Created page with "'''Solution: D''' <math>1.07^n=4</math> so <math>n \ln (1.07)=\ln (4)</math> so <math>n=\ln (4) / \ln (1.07)=20.48954</math> years. '''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
Solution: D
[math]1.07^n=4[/math] so [math]n \ln (1.07)=\ln (4)[/math] so [math]n=\ln (4) / \ln (1.07)=20.48954[/math] years.
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.