Worksheet for solving exponential equations with detailed solutions, including problems with bases like 4, 5, 9, 8, 3, 2, and 5.

We are given a worksheet with exponential equations to solve. The main section is Problem 2, which asks us to solve four equations and give answers to 3 significant figures.

Let’s solve each part one by one.

---

Problem 2a)

Equation:
\[
3^{2x} - 6(3^x) + 5 = 0
\]

This is a quadratic in form. Let:
\[
y = 3^x
\]
Then:
\[
3^{2x} = (3^x)^2 = y^2
\]

Substitute into the equation:
\[
y^2 - 6y + 5 = 0
\]

Factor:
\[
(y - 1)(y - 5) = 0
\Rightarrow y = 1 \quad \text{or} \quad y = 5
\]

Now substitute back \( y = 3^x \):

- If \( 3^x = 1 \), then \( x = \log_3(1) = 0 \)
- If \( 3^x = 5 \), then \( x = \log_3(5) = \frac{\ln 5}{\ln 3} \approx \frac{1.6094}{1.0986} \approx 1.465 \)

✔ Solutions:
\[
\boxed{x = 0.000} \quad \text{and} \quad \boxed{x = 1.47}
\]
*(to 3 significant figures)*

---

Problem 2b)

Equation:
\[
2^{2x} - 11(2^x) + 30 = 0
\]

Again, let:
\[
y = 2^x \Rightarrow 2^{2x} = y^2
\]

Substitute:
\[
y^2 - 11y + 30 = 0
\]

Factor:
\[
(y - 5)(y - 6) = 0
\Rightarrow y = 5 \quad \text{or} \quad y = 6
\]

Back-substitute:

- \( 2^x = 5 \Rightarrow x = \log_2(5) = \frac{\ln 5}{\ln 2} \approx \frac{1.6094}{0.6931} \approx 2.322 \)
- \( 2^x = 6 \Rightarrow x = \log_2(6) = \frac{\ln 6}{\ln 2} \approx \frac{1.7918}{0.6931} \approx 2.585 \)

✔ Solutions:
\[
\boxed{x = 2.32} \quad \text{and} \quad \boxed{x = 2.59}
\]
*(to 3 significant figures)*

---

Problem 2c)

Equation:
\[
5^{2x} - 4(5^x) - 32 = 0
\]

Let:
\[
y = 5^x \Rightarrow 5^{2x} = y^2
\]

Substitute:
\[
y^2 - 4y - 32 = 0
\]

Use quadratic formula:
\[
y = \frac{4 \pm \sqrt{(-4)^2 + 4 \cdot 32}}{2} = \frac{4 \pm \sqrt{16 + 128}}{2} = \frac{4 \pm \sqrt{144}}{2} = \frac{4 \pm 12}{2}
\]

So:
\[
y = \frac{4 + 12}{2} = 8 \quad \text{or} \quad y = \frac{4 - 12}{2} = -4
\]

But \( y = 5^x > 0 \), so discard \( y = -4 \).

Only valid solution: \( y = 8 \Rightarrow 5^x = 8 \Rightarrow x = \log_5(8) = \frac{\ln 8}{\ln 5} \approx \frac{2.0794}{1.6094} \approx 1.292 \)

✔ Solution:
\[
\boxed{x = 1.29}
\]
*(to 3 significant figures)*

---

Problem 2d)

Equation:
\[
3^{2x} + 2(3^x) - 15 = 0
\]

Let:
\[
y = 3^x \Rightarrow 3^{2x} = y^2
\]

Substitute:
\[
y^2 + 2y - 15 = 0
\]

Factor:
\[
(y + 5)(y - 3) = 0
\Rightarrow y = -5 \quad \text{or} \quad y = 3
\]

Discard \( y = -5 \) since \( 3^x > 0 \).

So \( y = 3 \Rightarrow 3^x = 3 \Rightarrow x = 1 \)

✔ Solution:
\[
\boxed{x = 1.00}
\]
*(to 3 significant figures — note that 1.00 emphasizes 3 sig figs)*

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## ✔ Final Answers (Problem 2 only, as requested):

a) \( x = 0.000 \), \( x = 1.47 \)
b) \( x = 2.32 \), \( x = 2.59 \)
c) \( x = 1.29 \)
d) \( x = 1.00 \)

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> Note: Problems i) and j) at the top are not part of Problem 2, and the instruction says “solve the problem” — assuming it refers to Problem 2 since it's explicitly numbered and has instructions for 3 sig figs. If you also want solutions for i) and j), let me know! But based on context, Problem 2 is the main task.