Worksheet featuring 22 logarithmic equations to solve, including natural logarithms and various bases.

Solving Logarithmic Equations Worksheet

We will solve each logarithmic equation step by step. Let's go through them one by one.

---

#### 1) \( 9 \log_9 v = 0 \)

1. Divide both sides by 9:
\[
\log_9 v = 0
\]
2. Recall that \(\log_b a = 0\) implies \(a = b^0 = 1\):
\[
v = 9^0 = 1
\]

Solution:
\[
\boxed{v = 1}
\]

---

#### 2) \( -\log_9 n = 1 \)

1. Multiply both sides by \(-1\):
\[
\log_9 n = -1
\]
2. Rewrite in exponential form:
\[
n = 9^{-1} = \frac{1}{9}
\]

Solution:
\[
\boxed{n = \frac{1}{9}}
\]

---

#### 3) \( -7 - 10 \log_6 r = -27 \)

1. Add 7 to both sides:
\[
-10 \log_6 r = -20
\]
2. Divide both sides by \(-10\):
\[
\log_6 r = 2
\]
3. Rewrite in exponential form:
\[
r = 6^2 = 36
\]

Solution:
\[
\boxed{r = 36}
\]

---

#### 4) \( 7 \log_5 x - 4 = 17 \)

1. Add 4 to both sides:
\[
7 \log_5 x = 21
\]
2. Divide both sides by 7:
\[
\log_5 x = 3
\]
3. Rewrite in exponential form:
\[
x = 5^3 = 125
\]

Solution:
\[
\boxed{x = 125}
\]

---

#### 5) \( -4 \log_6 (-r) = -4 \)

1. Divide both sides by \(-4\):
\[
\log_6 (-r) = 1
\]
2. Rewrite in exponential form:
\[
-r = 6^1 = 6
\]
3. Solve for \(r\):
\[
r = -6
\]

Solution:
\[
\boxed{r = -6}
\]

---

#### 6) \( -4 + \log_2 (-8p) = -3 \)

1. Add 4 to both sides:
\[
\log_2 (-8p) = 1
\]
2. Rewrite in exponential form:
\[
-8p = 2^1 = 2
\]
3. Solve for \(p\):
\[
p = \frac{2}{-8} = -\frac{1}{4}
\]

Solution:
\[
\boxed{p = -\frac{1}{4}}
\]

---

#### 7) \( 4 - 8 \log_7 (2x) = -28 \)

1. Subtract 4 from both sides:
\[
-8 \log_7 (2x) = -32
\]
2. Divide both sides by \(-8\):
\[
\log_7 (2x) = 4
\]
3. Rewrite in exponential form:
\[
2x = 7^4 = 2401
\]
4. Solve for \(x\):
\[
x = \frac{2401}{2} = 1200.5
\]

Solution:
\[
\boxed{x = 1200.5}
\]

---

#### 8) \( 6 + 3 \log_5 (k - 6) = 15 \)

1. Subtract 6 from both sides:
\[
3 \log_5 (k - 6) = 9
\]
2. Divide both sides by 3:
\[
\log_5 (k - 6) = 3
\]
3. Rewrite in exponential form:
\[
k - 6 = 5^3 = 125
\]
4. Solve for \(k\):
\[
k = 125 + 6 = 131
\]

Solution:
\[
\boxed{k = 131}
\]

---

#### 9) \( 9 \log_3 (-5r - 3) = 36 \)

1. Divide both sides by 9:
\[
\log_3 (-5r - 3) = 4
\]
2. Rewrite in exponential form:
\[
-5r - 3 = 3^4 = 81
\]
3. Solve for \(r\):
\[
-5r = 81 + 3 = 84
\]
\[
r = \frac{84}{-5} = -\frac{84}{5}
\]

Solution:
\[
\boxed{r = -\frac{84}{5}}
\]

---

#### 10) \( \log_6 (9 - 7x) - 7 = -6 \)

1. Add 7 to both sides:
\[
\log_6 (9 - 7x) = 1
\]
2. Rewrite in exponential form:
\[
9 - 7x = 6^1 = 6
\]
3. Solve for \(x\):
\[
-7x = 6 - 9 = -3
\]
\[
x = \frac{-3}{-7} = \frac{3}{7}
\]

Solution:
\[
\boxed{x = \frac{3}{7}}
\]

---

#### 11) \( 9 \log_6 (2a + 1) + 6 = 33 \)

1. Subtract 6 from both sides:
\[
9 \log_6 (2a + 1) = 27
\]
2. Divide both sides by 9:
\[
\log_6 (2a + 1) = 3
\]
3. Rewrite in exponential form:
\[
2a + 1 = 6^3 = 216
\]
4. Solve for \(a\):
\[
2a = 216 - 1 = 215
\]
\[
a = \frac{215}{2} = 107.5
\]

Solution:
\[
\boxed{a = 107.5}
\]

---

#### 12) \( -3 + 8 \log_9 (3x + 7) = 29 \)

1. Add 3 to both sides:
\[
8 \log_9 (3x + 7) = 32
\]
2. Divide both sides by 8:
\[
\log_9 (3x + 7) = 4
\]
3. Rewrite in exponential form:
\[
3x + 7 = 9^4 = 6561
\]
4. Solve for \(x\):
\[
3x = 6561 - 7 = 6554
\]
\[
x = \frac{6554}{3}
\]

Solution:
\[
\boxed{x = \frac{6554}{3}}
\]

---

#### 13) \( \log_{15} (4 - p) = \log_{15} (-2p + 2) \)

1. Since the bases are the same, set the arguments equal:
\[
4 - p = -2p + 2
\]
2. Solve for \(p\):
\[
4 - 2 = -2p + p
\]
\[
2 = -p
\]
\[
p = -2
\]

Solution:
\[
\boxed{p = -2}
\]

---

#### 14) \( \log_2 (-4x + 2) = \log_2 (5x + 2) \)

1. Since the bases are the same, set the arguments equal:
\[
-4x + 2 = 5x + 2
\]
2. Solve for \(x\):
\[
-4x - 5x = 2 - 2
\]
\[
-9x = 0
\]
\[
x = 0
\]

Solution:
\[
\boxed{x = 0}
\]

---

#### 15) \( \log_{20} (-3x - 1) = \log_{20} (-4x - 4) \)

1. Since the bases are the same, set the arguments equal:
\[
-3x - 1 = -4x - 4
\]
2. Solve for \(x\):
\[
-3x + 4x = -4 + 1
\]
\[
x = -3
\]

Solution:
\[
\boxed{x = -3}
\]

---

#### 16) \( \log (4v + 10) = \log (10 - 5v) \)

1. Since the bases are the same, set the arguments equal:
\[
4v + 10 = 10 - 5v
\]
2. Solve for \(v\):
\[
4v + 5v = 10 - 10
\]
\[
9v = 0
\]
\[
v = 0
\]

Solution:
\[
\boxed{v = 0}
\]

---

#### 17) \( \log_9 (-3x) - \log_9 10 = \log_9 13 \)

1. Use the logarithmic property \(\log_b a - \log_b c = \log_b \left(\frac{a}{c}\right)\):
\[
\log_9 \left(\frac{-3x}{10}\right) = \log_9 13
\]
2. Since the bases are the same, set the arguments equal:
\[
\frac{-3x}{10} = 13
\]
3. Solve for \(x\):
\[
-3x = 13 \cdot 10 = 130
\]
\[
x = \frac{130}{-3} = -\frac{130}{3}
\]

Solution:
\[
\boxed{x = -\frac{130}{3}}
\]

---

#### 18) \( \log_9 (-3x) - \log_9 3 = \log_9 35 \)

1. Use the logarithmic property \(\log_b a - \log_b c = \log_b \left(\frac{a}{c}\right)\):
\[
\log_9 \left(\frac{-3x}{3}\right) = \log_9 35
\]
2. Simplify the argument:
\[
\log_9 (-x) = \log_9 35
\]
3. Since the bases are the same, set the arguments equal:
\[
-x = 35
\]
4. Solve for \(x\):
\[
x = -35
\]

Solution:
\[
\boxed{x = -35}
\]

---

#### 19) \( \log_7 4x - \log_7 9 = 1 \)

1. Use the logarithmic property \(\log_b a - \log_b c = \log_b \left(\frac{a}{c}\right)\):
\[
\log_7 \left(\frac{4x}{9}\right) = 1
\]
2. Rewrite in exponential form:
\[
\frac{4x}{9} = 7^1 = 7
\]
3. Solve for \(x\):
\[
4x = 7 \cdot 9 = 63
\]
\[
x = \frac{63}{4}
\]

Solution:
\[
\boxed{x = \frac{63}{4}}
\]

---

#### 20) \( \log_7 9 + \log_7 (x + 3) = 2 \)

1. Use the logarithmic property \(\log_b a + \log_b c = \log_b (ac)\):
\[
\log_7 [9(x + 3)] = 2
\]
2. Rewrite in exponential form:
\[
9(x + 3) = 7^2 = 49
\]
3. Solve for \(x\):
\[
9x + 27 = 49
\]
\[
9x = 49 - 27 = 22
\]
\[
x = \frac{22}{9}
\]

Solution:
\[
\boxed{x = \frac{22}{9}}
\]

---

#### 21) \( \ln (x - 8) + \ln 3 = 5 \)

1. Use the logarithmic property \(\ln a + \ln b = \ln (ab)\):
\[
\ln [3(x - 8)] = 5
\]
2. Rewrite in exponential form:
\[
3(x - 8) = e^5
\]
3. Solve for \(x\):
\[
x - 8 = \frac{e^5}{3}
\]
\[
x = 8 + \frac{e^5}{3}
\]

Solution:
\[
\boxed{x = 8 + \frac{e^5}{3}}
\]

---

#### 22) \( \log (x + 4) + \log 3 = 2 \)

1. Use the logarithmic property \(\log a + \log b = \log (ab)\):
\[
\log [3(x + 4)] = 2
\]
2. Rewrite in exponential form:
\[
3(x + 4) = 10^2 = 100
\]
3. Solve for \(x\):
\[
x + 4 = \frac{100}{3}
\]
\[
x = \frac{100}{3} - 4 = \frac{100}{3} - \frac{12}{3} = \frac{88}{3}
\]

Solution:
\[
\boxed{x = \frac{88}{3}}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1) & \ v = 1 \\
2) & \ n = \frac{1}{9} \\
3) & \ r = 36 \\
4) & \ x = 125 \\
5) & \ r = -6 \\
6) & \ p = -\frac{1}{4} \\
7) & \ x = 1200.5 \\
8) & \ k = 131 \\
9) & \ r = -\frac{84}{5} \\
10) & \ x = \frac{3}{7} \\
11) & \ a = 107.5 \\
12) & \ x = \frac{6554}{3} \\
13) & \ p = -2 \\
14) & \ x = 0 \\
15) & \ x = -3 \\
16) & \ v = 0 \\
17) & \ x = -\frac{130}{3} \\
18) & \ x = -35 \\
19) & \ x = \frac{63}{4} \\
20) & \ x = \frac{22}{9} \\
21) & \ x = 8 + \frac{e^5}{3} \\
22) & \ x = \frac{88}{3}
\end{aligned}
}
\]