Algebra 2 homework worksheet on continuous compound interest with word problems and equations.

We are given a set of problems involving continuous compound interest, which uses the formula:

\[
A = Pe^{rt}
\]

Where:
- \( A \) = final amount
- \( P \) = principal (initial investment)
- \( r \) = annual interest rate (as a decimal)
- \( t \) = time in years
- \( e \approx 2.71828 \)

Also, there are exponential equations to solve (problems 11–14), using properties of exponents.

Let’s solve each problem one by one, carefully.

---

Problem 1:
Kim invests $P = \$4000$, $r = 3\% = 0.03$, $t = 4$ years. Find $A$.

\[
A = 4000 \cdot e^{0.03 \cdot 4} = 4000 \cdot e^{0.12}
\]

Compute $e^{0.12}$:
Using calculator: $e^{0.12} \approx 1.12749685$

So:
\[
A \approx 4000 \cdot 1.12749685 = 4509.9874 \approx \boxed{\$4510.00}
\]

(Usually round to nearest cent.)

Let me compute more precisely:
$4000 \times 1.12749685182 = 4509.98740728$ → \$4,509.99

Wait — check rounding:
Standard is to round to nearest cent → \$4,509.99

But many textbooks round to nearest dollar unless specified. The problem doesn’t specify, but money problems usually want cents.

Let’s keep two decimals: \$4,509.99

Hold on — let me double-check with high precision:

Using Python-style:
`exp(0.12) = 1.1274968515793757`
× 4000 = 4509.987406317503 → \$4,509.99

✔ Final: \$4,509.99

---

Problem 2:
Dash: $P = 10{,}000$, $r = 0.03$, $t = 8$

\[
A = 10000 \cdot e^{0.03 \cdot 8} = 10000 \cdot e^{0.24}
\]

$e^{0.24} \approx 1.27124915$

So:
\[
A \approx 10000 \cdot 1.27124915 = 12712.4915 \to \boxed{\$12,712.49}
\]

Check: exp(0.24) = 1.27124915032 → ×10000 = 12712.4915032 → \$12,712.49

---

Problem 3:
Ashleigh wants to double her money: $A = 2P$, $P = 5000$, $r = 0.04$. Find $t$.

Use:
\[
2P = P e^{0.04 t} \Rightarrow 2 = e^{0.04 t}
\]

Take natural log both sides:
\[
\ln 2 = 0.04 t \Rightarrow t = \frac{\ln 2}{0.04}
\]

$\ln 2 \approx 0.69314718$

So:
\[
t = \frac{0.69314718}{0.04} = 17.3286795 \approx \boxed{17.3} \text{ years}
\]

(Rounded to nearest tenth as instructed)

---

Problem 4:
Cyndie: $r = 0.02$, $t = 6$, $A = 1691.25$, find $P$.

\[
1691.25 = P e^{0.02 \cdot 6} = P e^{0.12}
\]

We already know $e^{0.12} \approx 1.12749685$

So:
\[
P = \frac{1691.25}{1.12749685} \approx ?
\]

Compute:
1691.25 ÷ 1.12749685 ≈ 1500.00 (let’s verify)

1.12749685 × 1500 = 1691.245275 → very close to 1691.25

So $P = \boxed{\$1,500.00}$

✔ Exact: because $1500 \cdot e^{0.12} = 1500 \cdot 1.12749685158 = 1691.245277$, rounds to 1691.25.

So answer: \$1,500.00

---

Problem 5:
Jenn: $P = 2150$, $r = 0.02$, $A = 2733.19$, find $t$.

\[
2733.19 = 2150 \cdot e^{0.02 t}
\Rightarrow \frac{2733.19}{2150} = e^{0.02 t}
\]

Compute left side:
2733.19 ÷ 2150 = 1.2712511628

So:
\[
e^{0.02 t} = 1.2712511628
\Rightarrow 0.02 t = \ln(1.2712511628)
\]

$\ln(1.2712511628) \approx ?$

We know earlier $e^{0.24} \approx 1.27124915$, so ln(1.27125116) ≈ 0.240001

So:
\[
t = \frac{0.240001}{0.02} = 12.00005 \approx \boxed{12} \text{ years}
\]

Check:
$2150 \cdot e^{0.02 \cdot 12} = 2150 \cdot e^{0.24} \approx 2150 \cdot 1.27124915 = 2733.1856725$ → rounds to \$2,733.19

✔ So answer: 12 years

---

Problem 6:
Damara: $P = 3500$, $r = 0.02$, $t = 5$

\[
A = 3500 \cdot e^{0.02 \cdot 5} = 3500 \cdot e^{0.1}
\]

$e^{0.1} \approx 1.105170918$

So:
\[
A = 3500 \cdot 1.105170918 = 3868.098213 \to \boxed{\$3,868.10}
\]

(Rounded to nearest cent)

---

Problem 7:
Kim: $r = 0.04$, $t = 3$, $A = 18039.95$, find $P$.

\[
18039.95 = P e^{0.04 \cdot 3} = P e^{0.12}
\]

Again, $e^{0.12} \approx 1.1274968516$

So:
\[
P = \frac{18039.95}{1.1274968516} \approx ?
\]

Compute:
18039.95 ÷ 1.1274968516 ≈ 16000.00? Let's test:

1.1274968516 × 16000 = 18039.9496256 → extremely close to 18039.95

So $P = \boxed{\$16,000.00}$

✔ Confirmed.

---

Problem 8:
Chelsea: $P = 7500$, $r = 0.05$, $A = 10643.01$, find $t$.

\[
10643.01 = 7500 \cdot e^{0.05 t}
\Rightarrow \frac{10643.01}{7500} = e^{0.05 t}
\]

Compute ratio:
10643.01 ÷ 7500 = 1.419068

Now take ln:
$\ln(1.419068) \approx ?$

Use approximation or calculator:
$\ln(1.419068) \approx 0.3500$? Let’s compute:

We know $e^{0.35} = \exp(0.35) \approx 1.4190675$ — yes!

So:
\[
0.05 t = 0.35 \Rightarrow t = \frac{0.35}{0.05} = 7
\]

✔ So answer: 7 years

Check:
$7500 \cdot e^{0.05 \cdot 7} = 7500 \cdot e^{0.35} \approx 7500 \cdot 1.4190675 = 10643.00625$ → rounds to \$10,643.01

Perfect.

---

Problem 9:
Dash: $P = 4125$, $t = 5$, $A = 4193.89$, find $r$.

\[
4193.89 = 4125 \cdot e^{5r}
\Rightarrow \frac{4193.89}{4125} = e^{5r}
\]

Compute ratio:
4193.89 ÷ 4125 = 1.0167

More precisely:
4193.89 / 4125 = 1.016699... ≈ 1.0167

Now:
\[
5r = \ln(1.016699) \approx ?
\]

Use approximation: for small x, $\ln(1+x) \approx x - x^2/2$, but better use calculator.

$\ln(1.016699) \approx 0.016555$

Then:
\[
r = \frac{0.016555}{5} = 0.003311 \approx 0.3311\%
\]

But that seems too low — wait, check numbers again:

Is $A = 4193.89$, $P = 4125$, difference = 68.89 over 5 years → ~1.67% total growth, so ~0.33% per year? That seems plausible for continuous compounding.

But let me compute more accurately:

4193.89 / 4125 =
4193.89 ÷ 4125 =
Let me do long division:

4125 × 1.0167 = 4125 + 4125×0.0167 = 4125 + 68.8875 = 4193.8875 → matches 4193.89

So ratio = 1.0167 exactly enough.

Now ln(1.0167):

Using calculator: ln(1.0167) = 0.016556

Then r = 0.016556 / 5 = 0.0033112 → 0.33112%

But interest rates are usually expressed as percent, so 0.331%? That seems unusually low — maybe I misread the problem.

Wait — problem says: “Dash puts $4125 into an account… now has $4193.89” after 5 years.

Yes, that’s only ~1.67% total gain → ~0.33% annual continuous rate.

But maybe they expect answer as decimal or percent?

The question: “What is the interest rate?” — likely as percent.

So answer: 0.331% (rounded to 3 decimal places) or 0.33%?

Let’s see if exact value gives nice number:

Suppose r = 0.0033 = 0.33%, then:

$e^{5×0.0033} = e^{0.0165} ≈ 1.016636$, ×4125 = 4193.62 — a bit low.

r = 0.00331 → 5r = 0.01655 → e^0.01655 ≈ 1.016685 → ×4125 = 4193.82

r = 0.003312 → 5r = 0.01656 → e^0.01656 ≈ 1.016699 → ×4125 = 4193.89 ✔

So r = 0.003312 = 0.3312%

Rounded to nearest hundredth of a percent: 0.33%

But maybe they want decimal form: 0.00331

However, looking at other problems, rates are given as percentages (3%, 4.5%, etc.), so answer should be in percent.

Let’s give: 0.33% (rounded to nearest hundredth of a percent) — but the problem doesn’t specify rounding.

Alternatively, maybe I made a mistake: Could it be 3.31%? Let's test:

If r = 0.0331 (3.31%), then 5r = 0.1655, e^0.1655 ≈ 1.1800, ×4125 ≈ 4867 — too big.

So no, it's definitely ~0.33%.

I’ll go with 0.331% (to 3 decimal places) or as decimal 0.00331.

But let’s hold and come back — maybe later problems clarify.

---

Problem 10:
Ashleigh: $r = 0.045$, $t = 10$, $A = 3567.91$, find $P$.

\[
3567.91 = P e^{0.045 \cdot 10} = P e^{0.45}
\]

Compute $e^{0.45} \approx ?$

$e^{0.45} = \exp(0.45) \approx 1.568312185$

So:
\[
P = \frac{3567.91}{1.568312185} \approx ?
\]

Compute:
3567.91 ÷ 1.568312185 ≈ 2275.00?

Check: 1.568312185 × 2275 = ?

1.568312185 × 2000 = 3136.62437
1.568312185 × 275 = 431.2858509
Sum = 3567.91022 → matches 3567.91

So $P = \boxed{\$2,275.00}$

✔ Perfect.

---

Now the exponential equations:

11) $3^{-b} = 3^{-3b}$

Since bases are same and positive ≠ 1, exponents must be equal:

\[
-b = -3b \Rightarrow -b + 3b = 0 \Rightarrow 2b = 0 \Rightarrow b = 0
\]

✔ Answer: 0

12) $2^{3n} = \frac{1}{64}$

Note: $64 = 2^6$, so $\frac{1}{64} = 2^{-6}$

Thus:
\[
2^{3n} = 2^{-6} \Rightarrow 3n = -6 \Rightarrow n = -2
\]

✔ Answer: -2

13) $4^{-m} = 4^{m - 3}$

Same base ⇒ exponents equal:

\[
-m = m - 3 \Rightarrow -m - m = -3 \Rightarrow -2m = -3 \Rightarrow m = \frac{3}{2}
\]

✔ Answer: $\frac{3}{2}$ or 1.5

14) $\left(\frac{1}{6}\right)^{-k} = \frac{1}{36}$

Recall: $\left(\frac{1}{6}\right)^{-k} = 6^{k}$

And $\frac{1}{36} = 6^{-2}$

So:
\[
6^{k} = 6^{-2} \Rightarrow k = -2
\]

✔ Answer: -2

---

Now go back to Problem 9 — let’s express rate as percent rounded to nearest hundredth:

We had $r = 0.003312 = 0.3312\%$

Rounded to nearest hundredth of a percent → 0.33%

But sometimes financial problems expect rate as decimal (like 0.0033). However, all given rates in problems are in percent (3%, 4%, etc.), so answer should be in percent.

I’ll write 0.33%

But let me verify once more with full precision:

A = 4193.89
P = 4125
t = 5

r = (1/t) * ln(A/P) = (1/5) * ln(4193.89/4125)
= 0.2 * ln(1.0166990279)
ln(1.0166990279) = 0.016555999
×0.2 = 0.0033112 → 0.33112%

So to nearest hundredth of a percent: 0.33%
To nearest thousandth of a percent: 0.331%

Since other answers are exact (like 12, 7, 1.5), and this is approximate, I think they expect 0.33% or maybe as decimal 0.00331.

But looking at problem wording: “What is the interest rate?” — in context of earlier problems, they used “3%", so answer should be in percent.

I’ll put 0.33%

However, let me check if perhaps the amount was misread: $4125 → $4193.89 — yes, that’s correct.

Okay, final list:

1) \$4,509.99
2) \$12,712.49
3) 17.3 years
4) \$1,500.00
5) 12 years
6) \$3,868.10
7) \$16,000.00
8) 7 years
9) 0.33%
10) \$2,275.00
11) 0
12) -2
13) 3/2
14) -2

Now format Final Answer as requested: only the final answers, but since there are 14 items, we need to list them clearly.

The instruction says:
Final Answer: [Provide only the final correct answer] — but there are multiple questions.

Given the structure, likely they want all answers listed in order.

I will output the 14 answers in a clean list.

Final Answer:
1) \$4,509.99
2) \$12,712.49
3) 17.3
4) \$1,500.00
5) 12
6) \$3,868.10
7) \$16,000.00
8) 7
9) 0.33%
10) \$2,275.00
11) 0
12) -2
13) \frac{3}{2}
14) -2