Let’s break these problems down step by step.

Problem 1:

Formula:
$A = P(1 + r)^t$

Given:

• $A = 2600$
• $P = 2200$
• $t = 5$

Solve for $r$:

1. Rearrange the formula to isolate $(1 + r)$:
   $1 + r = \left(\frac{A}{P}\right)^{\frac{1}{t}}$

2. Substitute values:
   $1 + r = \left(\frac{2600}{2200}\right)^{\frac{1}{5}}$

3. Simplify: $1 + r = (1.1818)^{0.2}$

   Using a calculator: $1 + r \approx 1.0339$

4. Solve for $r$:
   $r = 1.0339 - 1 = 0.0339$

5. Convert $r$ to a percentage:
   $r \approx 3.39\%$

Problem 2:

Formula:
$A = P(1 + r)^t$

Given:

• $A = 45,000$
• $P = 30,000$
• $r = 0.07$

Solve for $t$:

1. Rearrange the formula to isolate $t$:
   $t = \frac{\log{\left(\frac{A}{P}\right)}}{\log{(1 + r)}}$

2. Substitute values:
   $t = \frac{\log{\left(\frac{45,000}{30,000}\right)}}{\log{(1.07)}}$

3. Simplify:
   $t = \frac{\log{1.5}}{\log{1.07}}$

   Using a calculator: $t = \frac{0.1761}{0.0296} \approx 5.95$

4. Round to the nearest whole number:
   $t \approx 6 \, \text{years}$

Problem 3:

Formula for simple interest:
$I = P \cdot r \cdot t$

Given:

• $P = 30,000$
• $r = 0.068$ (6.8% annual interest)
• $t = \frac{5}{12}$ (5 months = $\frac{5}{12}$ of a year)

Solve for $I$:

1. Substitute values:
   $I = 30,000 \cdot 0.068 \cdot \frac{5}{12}$

2. Simplify:
   $I = 30,000 \cdot 0.068 \cdot 0.4167$

   Using a calculator: $I = 850.02$

3. Round to the nearest cent:
   $I \approx 850.02$

Final Results:

1. Problem 1: $r \approx 3.39\%$
2. Problem 2: $t \approx 6 \, \text{years}$
3. Problem 3: Interest due: $\$850.02$

Would you like a deeper explanation or further assistance?

Related Questions:

1. How do we derive the compound interest formula?
2. Why is $t$ in years for these calculations?
3. What adjustments are needed for monthly compounding?
4. How does simple interest differ from compound interest?
5. What is the logarithmic rule used to isolate $t$?

Tip:

Always double-check your calculations with a calculator, especially for compound interest problems involving exponents and logarithms.