Question

several years ago, frederick and gail each bought a boat on the same day. since then, the value of their respective boats has been decreasing.
fredericks boat
function ( f ) described below can be used to determine the value of fredericks boat.
( f(x) = 70000b^x )
where
( x ): time elapsed, in years, since the boats were bought
( f(x) ): value of fredericks boat, in dollars

• exactly 2 years after frederick bought his boat, its value was $60543.
• to the nearest cent, the value of fredericks boat today is $45289.31.

gails boat
function ( g ) described below can be used to determine the value of gails boat.
( g(x) = a(0.94)^x )
where
( x ): time elapsed, in years, since the boats were bought
( g(x) ): value of gails boat, in dollars

• exactly 3 years after gail bought her boat, its value was $66446.72.

to the nearest cent, what is the value of gails boat today?

Explanation:

Step1: Find b for Frederick’s boat

Substitute $x=2$, $f(2)=60543$ into $f(x)=70000b^x$:
$60543 = 70000b^2$
$b^2 = \frac{60543}{70000} = 0.8649$
$b = \sqrt{0.8649} = 0.93$ (since value decreases, $b<1$)

Step2: Find current time x

Substitute current value $45289.31$ into $f(x)$:
$45289.31 = 70000(0.93)^x$
$\frac{45289.31}{70000} = (0.93)^x$
$0.64699 = (0.93)^x$
Take log: $x = \log_{0.93}(0.64699) = \frac{\ln(0.64699)}{\ln(0.93)} ≈ \frac{-0.435}{-0.0725} ≈ 6$ years

Step3: Find a for Gail’s boat

Substitute $x=3$, $g(3)=66446.72$ into $g(x)=a(0.94)^x$:
$66446.72 = a(0.94)^3$
$(0.94)^3 = 0.94×0.94×0.94 = 0.830584$
$a = \frac{66446.72}{0.830584} = 80000$

Step4: Calculate Gail’s current value

Current $x=6$, so $g(6)=80000(0.94)^6$
$(0.94)^6 ≈ (0.94^2)^3 = (0.8836)^3 ≈ 0.8836×0.8836×0.8836 ≈ 0.6857$
$g(6) ≈ 80000×0.6857 ≈ 54856.00$

Answer:

54856.00