Question

The Mendes family bought a new house 11 years ago for $100,000. The house is now worth $196,000. Assuming a steady rate of growth, what was the yearly rate of appreciation? Round your answer to the nearest tenth of a percent (1.2% etc)

Answer 1

so, let's see 11 years ago.. let's say our years "t" is 0, or the original starting price, so t = 0, and we know the house was 100000


`\bf \qquad \textit{Amount for Exponential change}\\\\ A=P(1\pm r)^t\qquad \begin{cases} A=\textit{accumulated amount}\to &100000\\ P=\textit{starting amount}\\ r=rate\\ t=\textit{elapsed period}\to &0\\ \end{cases} \\\\\\ 10000=P(1+r)^0\implies 10000=P\cdot 1\implies 100000=P\\\\ -------------------------------\\\\ meaning\implies \boxed{A=10000(1+r)^t}`

11 years later, t = 11, the price has appreciated to 196,000 what's the rate "r"?


`\bf A=10000(1+r)^t\ \begin{cases} A=196000\\ t=11 \end{cases}\implies 196000=100000(1+r)^(11) \\\\\\ \cfrac{196000}{100000}=(1+r)^(11)\implies \cfrac{49}{25}=(1+r)^(11)\implies \sqrt[11]{\cfrac{49}{25}}=1+r \\\\\\ \sqrt[11]{\cfrac{49}{25}}-1=r\implies 0.063087\approx r\qquad 0.063087\cdot 100\implies 6.3087 \\\\\\ 6.3\% \approx r`