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| rev | Admin | (Created page with "'''Solution: E''' Let P be the annual interest paid. The present value of John’s payments is <math>Pa_{\overline{X}|0.05}</math>. The present value of Karen’s payments is <math display = "block"> P(1.05)^{-X}\,a_{\underline{\infty}|0.05}=P(1.05)^{-X}\,/\,0.05. </math> Then, <math display = "block"> \begin{array}{l}{{P(1.05)^{-X}/{0.05=1.59P a_{\overline{x}|0.05}}}}\\ {{0.05=1.59-1.05^{-X}}}\\ {{\mathrm{~ln~39=2.59(1.05)^{-X}}}}\\ {{X=10.}}\end{array} </math>...") | Nov 18'23 at 10:40 | +499 |