Printable math worksheet featuring 14 problems for solving logarithmic equations.
Logarithmic Equations Worksheet with 14 algebra problems to solve.
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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 check domain: $3w^2 > 0$ → always true for $w \ne 0$, but since $w^2$ is used, both positive and negative values are valid.
✔ 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)
✔ Answer: No real solution
---
Set arguments equal:
$$
2q = 6q + 7 \\
-4q = 7 \\
q = -\frac{7}{4}
$$
Check domain:
- $2q > 0 \Rightarrow q > 0$
- $6q + 7 > 0 \Rightarrow q > -\frac{7}{6}$
But $q = -1.75 < 0$ → not valid
✔ 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: $4w^2 > 0$ → yes ✔
✔ 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:
- $8r = 8 \cdot \frac{7}{3} = \frac{56}{3} > 0$ ✔
- $5r + 7 = \frac{35}{3} + 7 = \frac{56}{3} > 0$ ✔
✔ Answer: $r = \frac{7}{3}$
---
Set arguments equal:
$$
7p + 6 = 5p - 9 \\
7p - 5p = -9 - 6 \\
2p = -15 \\
p = -\frac{15}{2}
$$
Check domain:
- $7p + 6 = 7(-7.5) + 6 = -52.5 + 6 = -46.5 < 0$ ✘
- $5p - 9 = -37.5 - 9 = -46.5 < 0$ ✘
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 with graphs!
---
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 check domain: $3w^2 > 0$ → always true for $w \ne 0$, but since $w^2$ is used, both positive and negative values are valid.
✔ 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)
✔ 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:
- $2q > 0 \Rightarrow q > 0$
- $6q + 7 > 0 \Rightarrow q > -\frac{7}{6}$
But $q = -1.75 < 0$ → not valid
✔ Answer: No solution
---
8) $\log_7 (5x + 3) = \log_7 (9x - 9)$
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: $4w^2 > 0$ → yes ✔
✔ 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:
- $8r = 8 \cdot \frac{7}{3} = \frac{56}{3} > 0$ ✔
- $5r + 7 = \frac{35}{3} + 7 = \frac{56}{3} > 0$ ✔
✔ Answer: $r = \frac{7}{3}$
---
14) $\log_3 (7p + 6) = \log_3 (5p - 9)$
Set arguments equal:
$$
7p + 6 = 5p - 9 \\
7p - 5p = -9 - 6 \\
2p = -15 \\
p = -\frac{15}{2}
$$
Check domain:
- $7p + 6 = 7(-7.5) + 6 = -52.5 + 6 = -46.5 < 0$ ✘
- $5p - 9 = -37.5 - 9 = -46.5 < 0$ ✘
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 with graphs!
Parent Tip: Review the logic above to help your child master the concept of math equations worksheet.