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Logarithmic Equations worksheet with 14 problems for solving logarithmic equations.

A worksheet titled "Logarithmic Equations" with 14 problems for solving logarithmic equations, including expressions like log 9q = log (4q - 6) and log₄ (5x + 10) = log₄ (9x + 3), designed for math practice.

A worksheet titled "Logarithmic Equations" with 14 problems for solving logarithmic equations, including expressions like log 9q = log (4q - 6) and log₄ (5x + 10) = log₄ (9x + 3), designed for math practice.

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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. We'll use logarithmic properties such as:

- Logarithmic Equality Rule: If $\log_b A = \log_b B$, then $A = B$ (for $A > 0$, $B > 0$).
- Product Rule: $\log_b A + \log_b B = \log_b (AB)$
- Quotient Rule: $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$
- Power Rule: $\log_b (A^n) = n \log_b A$
- Change of base: Not needed here, but useful in general.

---

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



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

Check domain:
We must ensure arguments of log are positive:
- $9q = 9(-6/5) = -54/5 < 0$ → Invalid
- $4q - 6 = 4(-6/5) - 6 = -24/5 - 30/5 = -54/5 < 0$

Both negative → 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:
- $5x + 10 = 5(7/4) + 10 = 35/4 + 40/4 = 75/4 > 0$
- $9x + 3 = 9(7/4) + 3 = 63/4 + 12/4 = 75/4 > 0$

Valid

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

---

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



Use product rule:
$$
\log_3 (s \cdot 3) = 2 \\
\log_3 (3s) = 2 \\
3s = 3^2 = 9 \\
s = 3
$$

Check: $s = 3 > 0$ → Valid

Answer: $s = 3$

---

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



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

Check: $q = 2/3 > 0$ → Valid

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

---

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



Combine 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 check domain: $3w^2 > 0$ → always true for $w \ne 0$, but since $w^2$ is involved, both values valid unless restricted.

But we need to check if $3w^2 > 0$: yes, even for negative $w$, $w^2 > 0$. So both 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 square)

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:
- $2q = 2(-7/4) = -7/2 < 0$ → invalid
- $6q + 7 = 6(-7/4) + 7 = -42/4 + 28/4 = -14/4 < 0$ → invalid

Both arguments negative → 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:
- $5x + 3 = 15 + 3 = 18 > 0$
- $9x - 9 = 27 - 9 = 18 > 0$

Valid

Answer: $x = 3$

---

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



Use product rule:
$$
\log_2 (4n) = 2 \\
4n = 2^2 = 4 \\
n = 1
$$

Check: $n = 1 > 0$ → valid

Answer: $n = 1$

---

10) $\log_8 n + \log_8 8 = \log_8 10$



$$
\log_8 (8n) = \log_8 10 \\
8n = 10 \\
n = \frac{10}{8} = \frac{5}{4}
$$

Check: $n = 5/4 > 0$ → valid

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

---

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



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

Check domain: $4w^2 > 0$ → true for all $w \ne 0$, and $w^2 = 1/8 > 0$ → valid

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:
- $8r = 8(7/3) = 56/3 > 0$
- $5r + 7 = 35/3 + 21/3 = 56/3 > 0$

Valid

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

---

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



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

Check domain:
- $7p + 6 = 7(-15/2) + 6 = -105/2 + 12/2 = -93/2 < 0$ → invalid
- $5p - 9 = 5(-15/2) - 9 = -75/2 - 18/2 = -93/2 < 0$ → invalid

Both negative → 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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Parent Tip: Review the logic above to help your child master the concept of solving exponential equations using logarithms worksheet.
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