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[tex]\begin{gathered} \text{ Given} \\ A=290250 \\ r=0.114\text{ converted from percentage} \\ n=12,\text{ since it is compounded monthly (12 months in a year)} \\ A=P(1+\frac{r}{n})^{nt}\rightarrow\text{Compound Interest Formula} \\ \text{ A.) Substitute all the following values, that should leave us with} \\ \text{ P and t as variables} \\ A=P(1+\frac{r}{n})^{nt} \\ 290250=P(1+\frac{0.114}{12})^{12t}\rightarrow290250=P(1.0095)^{12t} \\ \text{ B.) Solve the equation for P} \\ \text{ We isolate P on the left hand side and we get} \\ 290250=P(1.0095)^{12t} \\ P(1.0095)^{12t}=290250 \\ \frac{P\cancel{(1.0095)^{12t}}}{\cancel{(1.0095)^{12t}}}=\frac{290250}{(1.0095)^{12t}} \\ P=\frac{290250}{(1.0095)^{12t}} \\ \text{ C.) Solve for t} \\ 290250=P(1.0095)^{12t} \\ P(1.0095)^{12t}=290250 \\ \frac{\cancel{P}(1.0095)^{12t}}{\cancel{P}}=\frac{290250}{P} \\ 1.0095^{12t}=\frac{290250}{P}\text{ get the natural logarithm of both sides} \\ \ln (1.0095^{12t})=\ln \frac{290250}{P} \\ 12t\cdot\ln 1.0095=\ln \frac{290250}{P},\text{ after this divide both sides by 12 and ln (1.0095)} \\ \text{ to isolate }t \\ \frac{\cancel{12}t\cdot\cancel{\ln (1.0095)}}{\cancel{12\cdot\ln(1.0095)}}=\frac{\ln\frac{290250}{P}}{12\cdot\ln(1.0095)} \\ t=\frac{\ln\frac{290250}{P}}{12\cdot\ln(1.0095)} \\ \text{ D.) Using the equation that we have on B, just substitute the values} \\ \text{ with }t=17 \\ P=\frac{290250}{(1.0095)^{12t}} \\ P=\frac{290250}{(1.0095)^{12(17)}} \\ \text{ Input this in a calculator and we get} \\ P=\$42177.64 \\ \text{ E.) Using the equation that have on C, substitute the values with} \\ P=135000 \\ t=\frac{\ln\frac{290250}{P}}{12\cdot\ln(1.0095)} \\ t=\frac{\ln \frac{290250}{135000}}{12\cdot\ln (1.0095)} \\ \text{ Input this in a calculator and we get} \\ t=6.75\text{ years},\text{ rounded to a hundredth} \end{gathered}[/tex]
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