Question

A home is valued at $236,500 in 2012. In 2017, the home is worth $305,700. Assume the home's value is increasing exponentially. a. Construct a function V(t) which models the value of the home as a function of years since 2012. b. If the model holds, what year will the home value be worth $400,000?

Exponential models
ANSWERED
asked 2020-11-08
A home is valued at $236,500 in 2012. In 2017, the home is worth $305,700. Assume the home's value is increasing exponentially.
a. Construct a function V(t) which models the value of the home as a function of years since 2012.
b. If the model holds, what year will the home value be worth $400,000?

Answers (1)

2020-11-09
Consider the following exponential function of value of the home:
\(\displaystyle{V}{\left({t}\right)}={a}{b}^{{t}}\)
Where,
V(t)=Value of home after t years
t=number of years from 2012
a and b are constants
In 2012, value of home is $236,500
\(\displaystyle{a}{b}^{{0}}={236500}\)
\(\displaystyle{a}={236500}\)
In 2017, value of home is $305,700
\(\displaystyle{t}={2017}-{2012}\)
\(\displaystyle={5}\)
\(\displaystyle{236500}\times{b}^{{5}}={305700}\)
\(\displaystyle{b}^{{5}}={\frac{{{305700}}}{{{236500}}}}\)
\(\displaystyle{b}^{{5}}={1.2926}\)
\(\displaystyle{b}=\sqrt{{{5}}}{\left\lbrace{1.2926}\right\rbrace}\)
\(\displaystyle{b}={1.05267}\)
Hence, the exponential function is \(\displaystyle{V}{\left({t}\right)}={236500}\times{\left({1.05267}\right)}^{{t}}={400000}\)
\(\displaystyle{\left({1.05267}\right)}^{{t}}={\frac{{{400000}}}{{{236500}}}}\)
\(\displaystyle{\left({1.05267}\right)}^{{t}}={1.691332}\)
Take logarithm on both sides of the equation:
\(\displaystyle{t}\times{\log{{\left({1.05267}\right)}}}={\log{{\left({1.691332}\right)}}}\)
\(\displaystyle{t}={\frac{{{\log{{\left({1.691332}\right)}}}}}{{{\log{{\left({1.05267}\right)}}}}}}\)
\(\displaystyle={\frac{{{0.228229}}}{{{0.02229}}}}\)
\(\displaystyle={10.23}\)
\(\displaystyle\approx{10}\)
Hence, after 10 years value of home will become $400,000.
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