Question

4 a savings bond is initially bought for $250. it doubles in value every decade.
a. complete the table.

decades since bond is bought	dollar value of bond
1
2
3
d

b. how many decades does it take before the bond is worth more than $10,000?
c. write an equation relating v, the value of the bond, to d, the number of decades since the bond was bought.
5 a sea turtle population p is modeled by the equation $p = 400 \cdot \left(\frac{5}{4}\
ight)^y$ where y is the number of years since the population was first measured.
a. how many turtles are in the population when it is first measured?
where do you see this in the equation?
b. is the population increasing or decreasing?
how can you tell from the equation?
c. when will the turtle population reach 700?
explain how you know.

Explanation:

---

Problem 4

Step1: Fill initial bond value

At $d=0$, value = $\$250$

Step2: Calculate value at 1 decade

Double initial value: $250 \times 2 = \$500$

Step3: Calculate value at 2 decades

Double 1-decade value: $500 \times 2 = \$1000$

Step4: Calculate value at 3 decades

Double 2-decade value: $1000 \times 2 = \$2000$

Step5: Generalize for d decades

Value doubles each decade: $v = 250 \times 2^d$

Step6: Solve for $v > 10000$

Set up inequality: $250 \times 2^d > 10000$
Divide both sides by 250: $2^d > 40$
Use logarithms: $d > \frac{\ln(40)}{\ln(2)} \approx 5.32$
Round up to whole decades: $d=6$

Step1: Find initial turtle population

At $y=0$, any number to 0 power = 1: $p = 400 \times (\frac{5}{4})^0 = 400 \times 1 = 400$

Step2: Analyze population trend

Growth factor is $\frac{5}{4} = 1.25$, which is greater than 1, so population increases.

Step3: Solve for $p=700$

Set up equation: $700 = 400 \times (\frac{5}{4})^y$
Divide both sides by 400: $\frac{7}{4} = (\frac{5}{4})^y$
Use logarithms: $y = \frac{\ln(\frac{7}{4})}{\ln(\frac{5}{4})} \approx \frac{0.5596}{0.2231} \approx 2.51$

Answer:

Part a

Decades Since Bond Is Bought	Dollar Value of Bond
1	$\$500$
2	$\$1000$
3	$\$2000$
$d$	$250 \times 2^d$

Part b

6 decades

Part c

$v = 250 \times 2^d$

---

Problem 5