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Logarithmic Equations worksheet with 14 problems for solving logarithmic equations.

Worksheet titled "Logarithmic Equations" with 14 problems to solve, including logarithmic expressions and equations, with spaces for name, teacher, score, and date at the top.

Worksheet titled "Logarithmic Equations" with 14 problems to solve, including logarithmic expressions and equations, with spaces for name, teacher, score, and date at the top.

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Show Answer Key & Explanations Step-by-step solution for: Algebra 2 Worksheets | Exponential and Logarithmic Functions ...
Let's solve each of the logarithmic equations step by step, using logarithmic properties and algebraic techniques.

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Logarithmic Properties Used:


1. If $\log_b A = \log_b B$, then $A = B$ (for $A > 0, B > 0$)
2. $\log_b A + \log_b B = \log_b (A \cdot B)$
3. $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$
4. $n \log_b A = \log_b (A^n)$
5. $\log_b A = c \Rightarrow A = b^c$

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Now, let's solve each equation:

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1) $\log 9q = \log (4q - 6)$



Since logs are equal and same base:
$$
9q = 4q - 6 \\
9q - 4q = -6 \\
5q = -6 \\
q = -\frac{6}{5}
$$

But check domain:
- $9q > 0 \Rightarrow q > 0$
- $4q - 6 > 0 \Rightarrow q > 1.5$

But $q = -1.2$ is not in domain → No solution

Answer: No solution

---

2) $\log_4 (5x + 10) = \log_4 (9x + 3)$



Set arguments equal:
$$
5x + 10 = 9x + 3 \\
10 - 3 = 9x - 5x \\
7 = 4x \\
x = \frac{7}{4}
$$

Check domain:
- $5x + 10 > 0$: $5(1.75) + 10 = 8.75 + 10 = 18.75 > 0$
- $9x + 3 > 0$: $9(1.75) + 3 = 15.75 + 3 = 18.75 > 0$

Answer: $x = \frac{7}{4}$

---

3) $\log_3 s + \log_3 3 = 2$



Use log addition:
$$
\log_3 (s \cdot 3) = 2 \\
\log_3 (3s) = 2 \\
3s = 3^2 = 9 \\
s = 3
$$

Check: $s > 0$ → yes

Answer: $s = 3$

---

4) $\log q + \log 6 = \log 4$



Assume base 10:
$$
\log (6q) = \log 4 \\
6q = 4 \\
q = \frac{2}{3}
$$

Check: $q > 0$ → yes

Answer: $q = \frac{2}{3}$

---

5) $\log_6 9 + \log_6 3w^2 = \log_6 7$



Add logs:
$$
\log_6 (9 \cdot 3w^2) = \log_6 7 \\
\log_6 (27w^2) = \log_6 7 \\
27w^2 = 7 \\
w^2 = \frac{7}{27} \\
w = \pm \sqrt{\frac{7}{27}} = \pm \frac{\sqrt{21}}{9}
$$

But since $w^2$ is inside log, we need $3w^2 > 0$ → always true for $w \ne 0$, but original log has $3w^2$, so no restriction on sign.

So both values are valid? Wait — actually, $w^2$ is fine, but let’s check if expression is defined.

Yes, as long as $w \in \mathbb{R}, w \ne 0$, but here $w^2$ is positive anyway.

So both solutions are acceptable.

Answer: $w = \pm \frac{\sqrt{21}}{9}$

---

6) $\log_3 (5p^2 + 180) = \log_3 65$



Set arguments equal:
$$
5p^2 + 180 = 65 \\
5p^2 = 65 - 180 = -115 \\
p^2 = -23
$$

No real solution (negative under square root)

Answer: No real solution

---

7) $\log_4 2q = \log_4 (6q + 7)$



Set arguments equal:
$$
2q = 6q + 7 \\
-4q = 7 \\
q = -\frac{7}{4}
$$

Check domain:
- $2q > 0 \Rightarrow q > 0$ → not satisfied
- $6q + 7 > 0$: $6(-1.75) + 7 = -10.5 + 7 = -3.5 < 0$ → invalid

Not in domain → No solution

Answer: No solution

---

8) $\log_4 (5x + 3) = \log_4 (9x - 9)$



Set arguments equal:
$$
5x + 3 = 9x - 9 \\
3 + 9 = 9x - 5x \\
12 = 4x \\
x = 3
$$

Check domain:
- $5x + 3 = 15 + 3 = 18 > 0$
- $9x - 9 = 27 - 9 = 18 > 0$

Answer: $x = 3$

---

9) $\log_2 n + \log_2 4 = 2$



Use log addition:
$$
\log_2 (4n) = 2 \\
4n = 2^2 = 4 \\
n = 1
$$

Check: $n > 0$ → yes

Answer: $n = 1$

---

10) $\log_6 n + \log_6 8 = \log_6 10$



Add logs:
$$
\log_6 (8n) = \log_6 10 \\
8n = 10 \\
n = \frac{10}{8} = \frac{5}{4}
$$

Check: $n > 0$ → yes

Answer: $n = \frac{5}{4}$

---

11) $\log_5 8 + \log_5 4w^2 = \log_5 4$



Add logs:
$$
\log_5 (8 \cdot 4w^2) = \log_5 4 \\
\log_5 (32w^2) = \log_5 4 \\
32w^2 = 4 \\
w^2 = \frac{4}{32} = \frac{1}{8} \\
w = \pm \frac{1}{2\sqrt{2}} = \pm \frac{\sqrt{2}}{4}
$$

Check: $4w^2 > 0$ → always true for $w \ne 0$, and here it's positive → okay

Answer: $w = \pm \frac{\sqrt{2}}{4}$

---

12) $\log_9 (9b^2 + 243) = \log_9 108$



Set arguments equal:
$$
9b^2 + 243 = 108 \\
9b^2 = 108 - 243 = -135 \\
b^2 = -15
$$

No real solution

Answer: No real solution

---

13) $\log_8 8r = \log_8 (5r + 7)$



Set arguments equal:
$$
8r = 5r + 7 \\
3r = 7 \\
r = \frac{7}{3}
$$

Check domain:
- $8r > 0$: $8 \cdot \frac{7}{3} > 0$
- $5r + 7 = 5 \cdot \frac{7}{3} + 7 = \frac{35}{3} + \frac{21}{3} = \frac{56}{3} > 0$

Answer: $r = \frac{7}{3}$

---

14) $\log_3 (7p + 6) = \log_3 (5p - 9)$



Set arguments equal:
$$
7p + 6 = 5p - 9 \\
7p - 5p = -9 - 6 \\
2p = -15 \\
p = -\frac{15}{2} = -7.5
$$

Check domain:
- $7p + 6 = 7(-7.5) + 6 = -52.5 + 6 = -46.5 < 0$
- $5p - 9 = 5(-7.5) - 9 = -37.5 - 9 = -46.5 < 0$

Not in domain → No solution

Answer: No solution

---

Final Answers Summary:



| Problem | Answer |
|--------|--------|
| 1 | No solution |
| 2 | $x = \frac{7}{4}$ |
| 3 | $s = 3$ |
| 4 | $q = \frac{2}{3}$ |
| 5 | $w = \pm \frac{\sqrt{21}}{9}$ |
| 6 | No real solution |
| 7 | No solution |
| 8 | $x = 3$ |
| 9 | $n = 1$ |
| 10 | $n = \frac{5}{4}$ |
| 11 | $w = \pm \frac{\sqrt{2}}{4}$ |
| 12 | No real solution |
| 13 | $r = \frac{7}{3}$ |
| 14 | No solution |

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