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.
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Step-by-step solution for: Algebra 2 Worksheets | Exponential and Logarithmic Functions ...
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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.
---
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. $\log_b A = c \Rightarrow A = b^c$
---
Now, let's solve each equation:
---
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
---
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: $x = \frac{7}{4}$
---
Use log addition:
$$
\log_3 (s \cdot 3) = 2 \\
\log_3 (3s) = 2 \\
3s = 3^2 = 9 \\
s = 3
$$
Check: $s > 0$ → yes ✔
✔ Answer: $s = 3$
---
Assume base 10:
$$
\log (6q) = \log 4 \\
6q = 4 \\
q = \frac{2}{3}
$$
Check: $q > 0$ → yes ✔
✔ Answer: $q = \frac{2}{3}$
---
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: $w = \pm \frac{\sqrt{21}}{9}$
---
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
---
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
---
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: $x = 3$
---
Use log addition:
$$
\log_2 (4n) = 2 \\
4n = 2^2 = 4 \\
n = 1
$$
Check: $n > 0$ → yes ✔
✔ Answer: $n = 1$
---
Add logs:
$$
\log_6 (8n) = \log_6 10 \\
8n = 10 \\
n = \frac{10}{8} = \frac{5}{4}
$$
Check: $n > 0$ → yes ✔
✔ Answer: $n = \frac{5}{4}$
---
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: $w = \pm \frac{\sqrt{2}}{4}$
---
Set arguments equal:
$$
9b^2 + 243 = 108 \\
9b^2 = 108 - 243 = -135 \\
b^2 = -15
$$
No real solution
✔ Answer: No real solution
---
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: $r = \frac{7}{3}$
---
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
---
| 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 |
---
Let me know if you'd like a PDF version or explanation in another format!
---
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. $\log_b A = c \Rightarrow A = b^c$
---
Now, let's solve each equation:
---
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 |
---
Let me know if you'd like a PDF version or explanation in another format!
Parent Tip: Review the logic above to help your child master the concept of solving linear equations worksheet algebra 2.