# In-Class Activity 3.1

### Question 1
Solve the following:

a) $$ 2^{x/3} = 4 $$

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<summary> Example: </summary>

Solve $3^{x/5}=7$.

| Explanation | Equation |
|-----------------------------------------------------------------------------|-------------------------------------------------|
| Start with the given equation. | $3^{x/5} = 7$ |
| Take the natural logarithm (ln) of both sides to eliminate the exponent. | $\ln(3^{x/5}) = \ln(7)$ |
| Apply the logarithmic power rule: $\ln(a^b) = b\ln(a)$. | $\frac{x}{5} \ln(3) = \ln(7)$ |
| Multiply both sides by $5$ to isolate $x$. | $x \ln(3) = 5 \ln(7)$ |
| Divide both sides by $\ln(3)$ to solve for $x$. | $x = \frac{5 \ln(7)}{\ln(3)}$ |
| Use a calculator to compute the logarithms and the division. | $x \approx \frac{5 \times 1.9459}{1.0986}$ |
| Final solution. | $x \approx 8.858$ |

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b) $$ 3^{2x-1} = \frac{1}{9} $$

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<summary> Example: </summary>

| Explanation | Equation |
|-----------------------------------------------------------------------------|-------------------------------------------------|
| Start with the given equation. | $4^{3x+2} = \frac{1}{16}$ |
| Express $\frac{1}{16}$ as $4^{-2}$ because $16 = 4^2$ and $\frac{1}{16}$ is the reciprocal of $16$. | $4^{3x+2} = 4^{-2}$ |
| Since the bases are the same, we can set the exponents equal to each other. | $3x + 2 = -2$ |
| Subtract $2$ from both sides to isolate the $3x$ term. | $3x + 2 - 2 = -2 - 2$ |
| Simplify the equation. | $3x = -4$ |
| Divide both sides by $3$ to solve for $x$. | $x = \frac{-4}{3}$ |
| Final solution. | $x = -\frac{4}{3}$ |

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c) $$ e^{0.5x} = 0.12 $$

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<summary> Example: </summary>

| Explanation | Equation |
|-----------------------------------------------------------------------------|-------------------------------------------------|
| Start with the given equation. | $e^{0.3x} = 0.53$ |
| Take the natural logarithm (ln) of both sides to eliminate the exponential. | $\ln(e^{0.3x}) = \ln(0.53)$ |
| Apply the logarithmic property: $\ln(e^y) = y$. | $0.3x = \ln(0.53)$ |
| Divide both sides by $0.3$ to isolate $x$. | $x = \frac{\ln(0.53)}{0.3}$ |
| Use a calculator to compute the natural logarithm and division. | $x \approx \frac{-0.6348}{0.3}$ |
| Final solution. | $x \approx -2.116$ |

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### Question 2
Solve the following:

a) $$ \log_2(3x) = 5 $$

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<summary> Example: </summary>

| Explanation | Equation |
|-----------------------------------------------------------------------------|-------------------------------------------------|
| Start with the given equation. | $\log_3(4x) = 7$ |
| Rewrite the logarithmic equation in its exponential form: $\log_b(a) = c \iff b^c = a$. | $4x = 3^7$ |
| Simplify $3^7$. | $4x = 2187$ |
| Divide both sides by 4 to isolate $x$. | $x = \frac{2187}{4}$ |
| Simplify the result. | $x = 546.75$ |

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b) $$ \log_{10}(2x-1) = -1 $$

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<summary> Example: </summary>

| Explanation | Equation |
|-----------------------------------------------------------------------------|-------------------------------------------------|
| Start with the given equation. | $\log(5x - 3) = -1$ |
| Rewrite the logarithmic equation in its exponential form: $\log_b(a) = c \iff 10^c = a$. | $5x - 3 = 10^{-1}$ |
| Simplify $10^{-1}$. | $5x - 3 = 0.1$ |
| Add $3$ to both sides to isolate the $5x$ term. | $5x = 0.1 + 3$ |
| Simplify the right-hand side. | $5x = 3.1$ |
| Divide both sides by $5$ to solve for $x$. | $x = \frac{3.1}{5}$ |
| Final solution. | $x = 0.62$ |

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c) $$ \ln(1-x) = 2 $$

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<summary> Example: </summary>

| Explanation | Equation |
|-----------------------------------------------------------------------------|-------------------------------------------------|
| Start with the given equation. | $\ln(3 - x) = 4$ |
| Rewrite the logarithmic equation in its exponential form: $\ln(a) = b \iff e^b = a$. | $3 - x = e^4$ |
| Subtract $3$ from both sides to isolate the $-x$ term. | $-x = e^4 - 3$ |
| Multiply both sides by $-1$ to solve for $x$. | $x = 3 - e^4$ |
| Use a calculator to approximate $e^4$. | $e^4 \approx 54.5982$ |
| Final solution after calculation. | $x \approx 3 - 54.5982 = -51.5982$ |
| Final rounded solution. | $x \approx -51.6$ |

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### Question 3
Provident Bank offers an investment that earns 1.64% compounded continuously.

a) If $10,000$ is invested, how much will it be worth in 3 years?

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**Skibidi Bank offers an investment that earns 5.63% interest, compounded continuously.**

**a) If $12,345 is invested, how much will it be worth in 5 years?**

### Solution:

We use the formula for continuous compounding:

$$ A = P e^{rt} $$

Where:
- $A$ = the amount of money after time $t$ (future value),
- $P$ = the initial investment ($12,345),
- $r$ = the annual interest rate (5.63% or 0.0563),
- $t$ = the time in years (5),
- $e$ = Euler's number (approximately 2.718).

### Step-by-step:

1. **Given values:**
- $P = 12345$
- $r = 0.0563$
- $t = 5$

2. **Plug into the formula:**

$$ A = 12345 \times e^{0.0563 \times 5} $$

3. **Calculate the exponent:**

$$ 0.0563 \times 5 = 0.2815 $$

4. **Calculate $e^{0.2815}$:**

$$ e^{0.2815} \approx 1.3252 $$

5. **Multiply by the initial investment:**

$$ A = 12345 \times 1.3252 \approx 16358.79 $$

### Answer:
The investment will be worth approximately **$16,358.79** after 5 years.

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b) How long will it take for the account to be worth $11,000$?

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<summary> Example: </summary>

**How long will it take for $12,345 to grow to $36,153?**

### Formula for continuous compounding:

$$ A = P e^{rt} $$

Where:
- $A$ = future value ($36,153),
- $P$ = initial investment ($12,345),
- $r$ = annual interest rate (5.63% or 0.0563),
- $t$ = time in years (this is what we're solving for),
- $e$ = Euler's number (approximately 2.718).

### Step-by-step solution:

| Explanation | Equation |
|--------------------------------------------------|---------------------------------------------|
| Start with the formula for continuous compounding. | $A = P e^{rt}$ |
| Substitute the known values: $A = 36153$, $P = 12345$, and $r = 0.0563$. | $36153 = 12345 \times e^{0.0563t}$ |
| Divide both sides by $12345$ to isolate the exponential term. | $\frac{36153}{12345} = e^{0.0563t}$ |
| Simplify the division. | $2.929 = e^{0.0563t}$ |
| Take the natural logarithm (ln) of both sides to eliminate $e$. | $\ln(2.929) = 0.0563t$ |
| Compute $\ln(2.929)$. | $1.0745 = 0.0563t$ |
| Divide both sides by $0.0563$ to solve for $t$. | $t = \frac{1.0745}{0.0563}$ |
| Final result. | $t \approx 19.08$ |

### Answer:
It will take approximately **19.08 years** for the account to grow to $36,153.

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### Question 4
An investor bought stock for $20,000. Five years later, the stock was sold for $30,000.
If interest is compounded continuously, what annual nominal rate of interest did the original investment earn?

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**Aaron Moosk bought stock for $420,000. Seven years later, the stock was sold for $666,000.**
**If interest is compounded continuously, what annual nominal rate of interest did the original investment earn?**

### Formula for continuous compounding:

$$ A = P e^{rt} $$

Where:
- $A$ = future value ($666,000),
- $P$ = initial investment ($420,000),
- $r$ = annual interest rate (this is what we're solving for),
- $t$ = time in years (7),
- $e$ = Euler's number (approximately 2.718).

### Step-by-step solution:

| Explanation | Equation |
|--------------------------------------------------|---------------------------------------------|
| Start with the formula for continuous compounding. | $A = P e^{rt}$ |
| Substitute the known values: $A = 666000$, $P = 420000$, and $t = 7$. | $666000 = 420000 \times e^{7r}$ |
| Divide both sides by $420000$ to isolate the exponential term. | $\frac{666000}{420000} = e^{7r}$ |
| Simplify the division. | $1.5857 = e^{7r}$ |
| Take the natural logarithm (ln) of both sides to eliminate $e$. | $\ln(1.5857) = 7r$ |
| Compute $\ln(1.5857)$. | $0.4602 = 7r$ |
| Divide both sides by 7 to solve for $r$. | $r = \frac{0.4602}{7}$ |
| Final result. | $r \approx 0.0657$ |

### Answer:
The annual nominal rate of interest is approximately **6.57%**.

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### Question 5
A note will pay $50,000 at maturity 5 years from now. How much should you be willing to pay for the note now if money is worth 6.4% compounded continuously?

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<summary> Example: </summary>

**A note will pay $40,000 at maturity 10 years from now. How much should you be willing to pay for the note now if money is worth 8.2% compounded continuously?**

### Formula for continuous compounding:

$$ A = P e^{rt} $$

Where:
- $A$ = future value ($40,000),
- $P$ = present value (this is what we're solving for),
- $r$ = annual interest rate (8.2% or 0.082),
- $t$ = time in years (10),
- $e$ = Euler's number (approximately 2.718).

### Step-by-step solution:

| Explanation | Equation |
|--------------------------------------------------|---------------------------------------------|
| Start with the formula for continuous compounding. | $A = P e^{rt}$ |
| Substitute the known values: $A = 40000$, $r = 0.082$, and $t = 10$. | $40000 = P \times e^{0.082 \times 10}$ |
| Simplify the exponent calculation. | $40000 = P \times e^{0.82}$ |
| Calculate $e^{0.82}$ using a calculator. | $e^{0.82} \approx 2.2715$ |
| Divide both sides by $2.2715$ to isolate $P$. | $P = \frac{40000}{2.2715}$ |
| Compute the result. | $P \approx 17,609.16$ |

### Answer:
You should be willing to pay approximately **$17,609.16** for the note now.

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