Free Printable Solving Logarithmic Equations Worksheets - Free Printable

Let's solve each logarithmic equation step by step and verify the provided answers. We'll go through the problems systematically, using logarithmic properties and algebraic manipulation.

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1) $ 9\log_9 v = 0 $

Step 1: Divide both sides by 9:
$$
\log_9 v = 0
$$

Step 2: Convert to exponential form:
$$
v = 9^0 = 1
$$

✔ Answer: $ \boxed{1} $ ✔️ (Matches given)

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2) $ -\log_9 n = 1 $

Step 1: Multiply both sides by -1:
$$
\log_9 n = -1
$$

Step 2: Convert to exponential form:
$$
n = 9^{-1} = \frac{1}{9}
$$

✔ Answer: $ \boxed{\frac{1}{9}} $ ✔️

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3) $ -7 - 10\log_6 r = -27 $

Step 1: Add 7 to both sides:
$$
-10\log_6 r = -20
$$

Step 2: Divide by -10:
$$
\log_6 r = 2
$$

Step 3: Convert:
$$
r = 6^2 = 36
$$

✔ Answer: $ \boxed{36} $ ✔️

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4) $ 7\log_5 x - 4 = 17 $

Step 1: Add 4:
$$
7\log_5 x = 21
$$

Step 2: Divide by 7:
$$
\log_5 x = 3
$$

Step 3: Convert:
$$
x = 5^3 = 125
$$

✔ Answer: $ \boxed{125} $ ✔️

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

Step 1: Divide both sides by -4:
$$
\log_6 (-r) = 1
$$

Step 2: Convert:
$$
-r = 6^1 = 6 \Rightarrow r = -6
$$

Note: The argument of log must be positive → $-r > 0$ → $r < 0$, so $r = -6$ is valid.

✔ Answer: $ \boxed{-6} $ ✔️

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

Step 1: Add 4:
$$
\log_2 (-8p) = 1
$$

Step 2: Convert:
$$
-8p = 2^1 = 2 \Rightarrow p = -\frac{2}{8} = -\frac{1}{4}
$$

Check: $-8p = -8(-1/4) = 2 > 0$, valid.

✔ Answer: $ \boxed{-\frac{1}{4}} $ ✔️

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7) $ 4 - 8\log_7 2x = -28 $

Step 1: Subtract 4:
$$
-8\log_7 2x = -32
$$

Step 2: Divide by -8:
$$
\log_7 2x = 4
$$

Step 3: Convert:
$$
2x = 7^4 = 2401 \Rightarrow x = \frac{2401}{2}
$$

✔ Answer: $ \boxed{\frac{2401}{2}} $ ✔️

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

Step 1: Subtract 6:
$$
3\log_5 (k - 6) = 9
$$

Step 2: Divide by 3:
$$
\log_5 (k - 6) = 3
$$

Step 3: Convert:
$$
k - 6 = 5^3 = 125 \Rightarrow k = 131
$$

✔ Answer: $ \boxed{131} $ ✔️

---

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

Step 1: Divide by 9:
$$
\log_3 (-5r - 3) = 4
$$

Step 2: Convert:
$$
-5r - 3 = 3^4 = 81
\Rightarrow -5r = 84 \Rightarrow r = -\frac{84}{5}
$$

Check: $-5r - 3 = -5(-84/5) - 3 = 84 - 3 = 81 > 0$, valid.

✔ Answer: $ \boxed{-\frac{84}{5}} $ ✔️

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

Step 1: Add 7:
$$
\log_6 (9 - 7x) = 1
$$

Step 2: Convert:
$$
9 - 7x = 6^1 = 6 \Rightarrow -7x = -3 \Rightarrow x = \frac{3}{7}
$$

✔ Answer: $ \boxed{\frac{3}{7}} $ ✔️

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

Step 1: Subtract 6:
$$
9\log_6 (2a + 1) = 27
$$

Step 2: Divide by 9:
$$
\log_6 (2a + 1) = 3
$$

Step 3: Convert:
$$
2a + 1 = 6^3 = 216 \Rightarrow 2a = 215 \Rightarrow a = \frac{215}{2}
$$

✔ Answer: $ \boxed{\frac{215}{2}} $ ✔️

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

Step 1: Add 3:
$$
8\log_9 (3x + 7) = 32
$$

Step 2: Divide by 8:
$$
\log_9 (3x + 7) = 4
$$

Step 3: Convert:
$$
3x + 7 = 9^4 = (3^2)^4 = 3^8 = 6561
\Rightarrow 3x = 6561 - 7 = 6554 \Rightarrow x = \frac{6554}{3}
$$

✔ Answer: $ \boxed{\frac{6554}{3}} $ ✔️

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

Since logs are equal with same base:
$$
4 - p = -2p + 2
\Rightarrow 4 - p = -2p + 2
\Rightarrow p = -2
$$

Check:
- Left: $4 - (-2) = 6 > 0$
- Right: $-2(-2) + 2 = 4 + 2 = 6 > 0$

Valid.

✔ Answer: $ \boxed{-2} $ ✔️

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

Set arguments equal:
$$
-4x + 2 = 5x + 2 \Rightarrow -9x = 0 \Rightarrow x = 0
$$

Check:
- $-4(0)+2 = 2 > 0$
- $5(0)+2 = 2 > 0$

Valid.

✔ Answer: $ \boxed{0} $ ✔️

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

Set arguments equal:
$$
-3x - 1 = -4x - 4 \Rightarrow x = -3
$$

Check:
- $-3(-3) - 1 = 9 - 1 = 8 > 0$
- $-4(-3) - 4 = 12 - 4 = 8 > 0$

Valid.

✔ Answer: $ \boxed{-3} $ ✔️

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

Assume base 10.

Set arguments equal:
$$
4v + 10 = 10 - 5v \Rightarrow 9v = 0 \Rightarrow v = 0
$$

Check:
- $4(0)+10 = 10 > 0$
- $10 - 5(0) = 10 > 0$

Valid.

✔ Answer: $ \boxed{0} $ ✔️

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

Use log property: $ \log_a A - \log_a B = \log_a(A/B) $
$$
\log_9\left(\frac{-3x}{10}\right) = \log_9 13
\Rightarrow \frac{-3x}{10} = 13
\Rightarrow -3x = 130 \Rightarrow x = -\frac{130}{3}
$$

Check: $-3x = -3(-130/3) = 130 > 0$, so $-3x > 0$, but log argument is $-3x$, which is positive → OK.

Wait: $\log_9(-3x)$ — the argument is $-3x$, which must be positive. Since $x = -130/3$, then $-3x = 130 > 0$, so valid.

✔ Answer: $ \boxed{-\frac{130}{3}} $ ✔️

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

Same method:
$$
\log_9\left(\frac{-3x}{3}\right) = \log_9 35
\Rightarrow \frac{-3x}{3} = 35 \Rightarrow -x = 35 \Rightarrow x = -35
$$

Check: $-3x = -3(-35) = 105 > 0$, so valid.

✔ Answer: $ \boxed{-35} $ ✔️

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19) $ \log_7 4x - \log_7 9 = 1 $

$$
\log_7\left(\frac{4x}{9}\right) = 1
\Rightarrow \frac{4x}{9} = 7^1 = 7
\Rightarrow 4x = 63 \Rightarrow x = \frac{63}{4}
$$

Check: $4x = 63 > 0$, valid.

✔ Answer: $ \boxed{\frac{63}{4}} $ ✔️

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

Use product rule:
$$
\log_7 [9(x + 3)] = 2
\Rightarrow 9(x + 3) = 7^2 = 49
\Rightarrow 9x + 27 = 49 \Rightarrow 9x = 22 \Rightarrow x = \frac{22}{9}
$$

Check: $x + 3 = \frac{22}{9} + \frac{27}{9} = \frac{49}{9} > 0$, valid.

✔ Answer: $ \boxed{\frac{22}{9}} $ ✔️

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

Use product rule:
$$
\ln[3(x - 8)] = 5
\Rightarrow 3(x - 8) = e^5
\Rightarrow 3x - 24 = e^5
\Rightarrow 3x = e^5 + 24
\Rightarrow x = \frac{e^5 + 24}{3}
$$

✔ Answer: $ \boxed{\frac{e^5 + 24}{3}} $ ✔️

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

Assume base 10.

$$
\log[3(x + 4)] = 2
\Rightarrow 3(x + 4) = 10^2 = 100
\Rightarrow 3x + 12 = 100 \Rightarrow 3x = 88 \Rightarrow x = \frac{88}{3}
$$

Check: $x + 4 = \frac{88}{3} + \frac{12}{3} = \frac{100}{3} > 0$, valid.

✔ Answer: $ \boxed{\frac{88}{3}} $ ✔️

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✔ Final Summary:

All solutions match the provided answers. Each problem was solved using:

- Logarithmic identities (product, quotient, power)
- Converting between logarithmic and exponential forms
- Solving linear equations
- Checking domain (arguments > 0)

The worksheet is correctly solved, and all answers are verified.

✔ All answers are correct!