Question

A young couple purchases their first new home in 2002 for $120,000. They sell it to move into bigger home in 2007 for $150,000. First, we will develop an exponential model for the value of the home. The model will have the form V(t)=V0ekt. Let t be years since 2002 and V(t) be the value of the home.

Answer 1

Answer:

The exponential model for the value of the home is $V(t)=120000e^{0.045t}$.

Step-by-step explanation:

According to the give information 2002 is the initial year and the value of the hom in 2002 is $120,000.

The model will have the form

$V(t)=V_0e^{kt}$

Where V₀ is initial value of home, k is a constant and t is number f years after 2002.

$V(t)=120000e^{kt}$

The value of home in 2007 is $150,000. Difference between 2007 and 2002 is 5 years. Therefore the value of function is 150000 at t=5.

$150000=120000e^{k(5)}$

$\frac{150000}{120000}=e^{5k}$

$\frac{5}{4}=e^{5k}$

Take ln both sides.

$ln(\frac{5}{4})=lne^{5k}$

$ln(\frac{5}{4})=5k$                    ($lne^a=a$)

$\frac{ln(\frac{5}{4})}{5}=k$

$k=0.04462871\approx 0.045$

Therefore exponential model for the value of the home is $V(t)=120000e^{0.045t}$.

Where t is number of years after 2002.

Answer 2

Answer:

$V(t)=120000*e^{(0.044)t}$

Step-by-step explanation:

If the given equation is $V(t)=V₀*e^{kt}$                .............(i)

Here from the question it is given that

V(t) = $150,000

V₀=$120,000

t= 2007-2002=5 years

e≅2.718

k=?

Now for the Value of k putting all values in equation (i)

$150000=120000*e^{k(5)}$

$e^{5(k)}=\frac{150000}{120000}$

$e^{5(k)}=\frac{5}{4}$

Now taking Natural log on both sides of the equation

㏑ (e^{5(k)})= ㏑\frac{5}{4}

as we Know that ln(e)=1

so

5k = 0.223

dividing both sides by 5 gives

k = $\frac{0.223}{5}$

k= 0.044

Now as we got the value of k we can form a general exponential equation which will be

$V(t)=V₀*e^{(0.044)t}$

here value of v₀ is 120000

so equation will be

$V(t)=120000*e^{(0.044)t}$