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(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}}") |
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'''Solution: B''' | |||
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 | 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"> | <math display="block"> | ||
n=\frac{\ln (200)-\ln \left(100(1+i)^2-50\right)}{\ln (1+i)} | n=\frac{\ln (200)-\ln \left(100(1+i)^2-50\right)}{\ln (1+i)}. | ||
</math> | </math> | ||
Latest revision as of 13:19, 26 November 2023
Solution: B
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)}.
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.