Solving Logarithm Equations Worksheet from Algebra 2, featuring 22 problems with solutions for practice.

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)

---

2) $ -\log_9 n = 1 $

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

Step 2: Exponential form:
$$
n = 9^{-1} = \frac{1}{9}
$$

✔ Answer: $ \boxed{\frac{1}{9}} $ ✔️

---

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: Exponential form:
$$
r = 6^2 = 36
$$

✔ Answer: $ \boxed{36} $ ✔️

---

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: Exponential form:
$$
x = 5^3 = 125
$$

✔ Answer: $ \boxed{125} $ ✔️

---

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

Note: The argument of log must be positive → $ -r > 0 \Rightarrow r < 0 $

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

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

✔ Answer: $ \boxed{-6} $ ✔️

---

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

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

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

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

✔ Answer: $ \boxed{-\frac{1}{4}} $ ✔️

---

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: Exponential form:
$$
2x = 7^4 = 2401 \Rightarrow x = \frac{2401}{2}
$$

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

---

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: Exponential form:
$$
k - 6 = 5^3 = 125 \Rightarrow k = 131
$$

✔ Answer: $ \boxed{131} $ ✔️

---

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

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

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

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

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

---

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

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

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

Check: $ 9 - 7(3/7) = 9 - 3 = 6 > 0 $ → valid

✔ Answer: $ \boxed{\frac{3}{7}} $ ✔️

---

11) $ 9\log_2 (2a + 1) + 6 = 33 $

Step 1: Subtract 6:
$$
9\log_2 (2a + 1) = 27
$$

Step 2: Divide by 9:
$$
\log_2 (2a + 1) = 3
$$

Step 3: Exponential form:
$$
2a + 1 = 2^3 = 8 \Rightarrow 2a = 7 \Rightarrow a = \frac{7}{2} = 3.5
$$

Wait — but answer is listed as $ \frac{215}{2} $? That’s 107.5, which is way off.

Let’s double-check:

Wait — maybe I misread the problem.

It says: $ 9\log_2 (2a + 1) + 6 = 33 $

We did:
- $ 9\log_2(...) = 27 $
- $ \log_2(...) = 3 $
- $ 2a + 1 = 8 $
- $ a = 3.5 = \frac{7}{2} $

But the answer given is $ \frac{215}{2} = 107.5 $. That can't be right unless the base or expression is different.

Wait — could it be $ \log_{10} $ or typo?

Wait — check again: Is it $ \log_2 $ or $ \log_{10} $?

No, it says $ \log_2 $. But let's suppose the answer is wrong? Or we made a mistake.

Wait — perhaps the equation is:
$$
9\log_2 (2a + 1) + 6 = 33
\Rightarrow 9\log_2(2a+1) = 27 \Rightarrow \log_2(2a+1) = 3 \Rightarrow 2a+1 = 8 \Rightarrow a = 3.5
$$

So correct answer should be $ \frac{7}{2} $, not $ \frac{215}{2} $.

But the worksheet says $ \boxed{\frac{215}{2}} $. That seems incorrect.

Wait — maybe the base is not 2?

Wait — no, it clearly says $ \log_2 $. Could it be $ \log_{10} $? No.

Alternatively, could it be $ 9\log_2 (2a + 1) = 33 - 6 = 27 $, yes.

Wait — unless the equation is $ 9\log_2 (2a + 1) + 6 = 33 $, which is what we have.

But $ \frac{215}{2} = 107.5 $. Try plugging in:

$ 2a + 1 = 2(107.5) + 1 = 215 + 1 = 216 $

Then $ \log_2(216) \approx \log_2(256) = 8 $, so $ \log_2(216) \approx 7.75 $, times 9 ≈ 69.75 + 6 = 75.75 ≠ 33.

So the given answer is wrong.

But wait — perhaps the base is not 2, but something else?

Wait — look at problem #11:
"11) $ 9\log_2 (2a + 1) + 6 = 33 $"

But maybe it's $ \log_{10} $? No, it says $ \log_2 $.

Alternatively, maybe it's $ \log_{10}(2a + 1) $? Let's try solving with $ \log_{10} $:

Suppose it was $ \log_{10} $, then:
$$
9\log_{10}(2a+1) = 27 \Rightarrow \log_{10}(2a+1) = 3 \Rightarrow 2a+1 = 10^3 = 1000 \Rightarrow 2a = 999 \Rightarrow a = 499.5 = \frac{999}{2}
$$

Still not $ \frac{215}{2} $

Wait — what if it's $ \log_2 (2a + 1) $, but the answer is actually $ \frac{7}{2} $? Then why does it say $ \frac{215}{2} $?

Wait — perhaps the problem is different.

Looking back: "11) $ 9\log_2 (2a + 1) + 6 = 33 $" — this is what's written.

But let's recompute:

$$
9\log_2(2a+1) = 27 \Rightarrow \log_2(2a+1) = 3 \Rightarrow 2a+1 = 8 \Rightarrow a = 3.5 = \frac{7}{2}
$$

But the box says $ \frac{215}{2} $. This is likely a mistake in the worksheet.

But wait — let's check problem #12:

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

Given answer: $ \frac{6554}{3} $

Let’s solve that one to see if pattern holds.

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

✔ Answer: $ \boxed{\frac{6554}{3}} $ ✔️ (Matches)

So that one is correct.

Back to #11: Our solution gives $ a = \frac{7}{2} $, but the worksheet says $ \frac{215}{2} $. So either:

- There's a typo in the problem or answer.
- Or maybe the base is different.

Wait — perhaps the base is not 2, but 10? Let's test:

If $ \log_{10}(2a+1) $, then:
$$
9\log_{10}(2a+1) = 27 \Rightarrow \log_{10}(2a+1) = 3 \Rightarrow 2a+1 = 1000 \Rightarrow a = 499.5
$$

Not matching.

What if it's $ \log_2(2a + 1) $, but the equation is different?

Wait — maybe it's $ 9\log_2 (2a + 1) + 6 = 33 $, and they want $ a = \frac{215}{2} $? Try plugging in:

$ a = 215/2 = 107.5 $

$ 2a + 1 = 215 + 1 = 216 $

$ \log_2(216) \approx \log_2(256) = 8 $, but $ \log_2(216) = \log_2(2^3 \cdot 27) = 3 + \log_2(27) \approx 3 + 4.75 = 7.75 $

Then $ 9 \times 7.75 = 69.75 $, plus 6 = 75.75 ≠ 33

So no, the answer $ \frac{215}{2} $ is wrong for this equation.

But wait — perhaps the equation is:

$ 9\log_2 (2a + 1) + 6 = 33 $

But maybe it's $ \log_{10} $? No.

Alternatively, could it be $ \log_2 (2a + 1) $, and the answer is $ \frac{7}{2} $, and the worksheet has a typo?

Yes — likely. So we’ll mark:

✘ Problem 11: Given answer $ \frac{215}{2} $ is incorrect. Correct answer is $ \boxed{\frac{7}{2}} $

But let’s continue.

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

Since logs are equal and same base, arguments must be equal:

$$
4 - p = -2p + 2
\Rightarrow -p + 2p = 2 - 4 \Rightarrow p = -2
$$

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

✔ Answer: $ \boxed{-2} $ ✔️

---

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

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

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

✔ Answer: $ \boxed{0} $ ✔️

---

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

Arguments equal:
$$
-3x - 1 = -4x - 4
\Rightarrow -3x + 4x = -4 + 1 \Rightarrow x = -3
$$

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) $

Assuming base 10.

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

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 rule: $ \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 $, so $ \frac{-3x}{10} = 130/10 = 13 $ → valid

Also, $ -3x = 130 > 0 $ → argument of log is positive → OK

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

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

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

Check: $ -3x = -3(-35) = 105 > 0 $, so log defined

✔ 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}} $ ✔️

---

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

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

Check: $ x + 3 = 22/9 + 27/9 = 49/9 > 0 $ → valid

✔ Answer: $ \boxed{\frac{22}{9}} $ ✔️

---

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

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

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

---

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 = 88/3 + 12/3 = 100/3 > 0 $ → valid

✔ Answer: $ \boxed{\frac{88}{3}} $ ✔️

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## ✔ Summary of Corrections

All answers match except Problem #11.

✘ Problem 11 Correction:

Given: $ 9\log_2(2a + 1) + 6 = 33 $

Solution:
- $ 9\log_2(2a+1) = 27 $
- $ \log_2(2a+1) = 3 $
- $ 2a + 1 = 8 $
- $ 2a = 7 $
- $ a = \frac{7}{2} $

But the worksheet says $ \frac{215}{2} $, which is incorrect.

Possibility: Maybe the base is not 2, but 10? Or the equation is different?

Wait — perhaps it's $ \log_{10} $? Let's try:

$ 9\log_{10}(2a+1) + 6 = 33 \Rightarrow \log_{10}(2a+1) = 3 \Rightarrow 2a+1 = 1000 \Rightarrow a = 499.5 $ — still not $ 215/2 $

Alternatively, could it be $ \log_2(2a + 1) $, but the answer is $ \frac{7}{2} $, and the printed answer is wrong?

Yes — likely a typo in the worksheet.

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## ✔ Final Answer Key:

| Problem | Solution |
|--------|----------|
| 1 | $ \boxed{1} $ |
| 2 | $ \boxed{\frac{1}{9}} $ |
| 3 | $ \boxed{36} $ |
| 4 | $ \boxed{125} $ |
| 5 | $ \boxed{-6} $ |
| 6 | $ \boxed{-\frac{1}{4}} $ |
| 7 | $ \boxed{\frac{2401}{2}} $ |
| 8 | $ \boxed{131} $ |
| 9 | $ \boxed{-\frac{84}{5}} $ |
| 10 | $ \boxed{\frac{3}{7}} $ |
| 11 | Correct: $ \boxed{\frac{7}{2}} $ (Worksheet says $ \frac{215}{2} $ — ✘ error) |
| 12 | $ \boxed{\frac{6554}{3}} $ |
| 13 | $ \boxed{-2} $ |
| 14 | $ \boxed{0} $ |
| 15 | $ \boxed{-3} $ |
| 16 | $ \boxed{0} $ |
| 17 | $ \boxed{-\frac{130}{3}} $ |
| 18 | $ \boxed{-35} $ |
| 19 | $ \boxed{\frac{63}{4}} $ |
| 20 | $ \boxed{\frac{22}{9}} $ |
| 21 | $ \boxed{\frac{e^5 + 24}{3}} $ |
| 22 | $ \boxed{\frac{88}{3}} $ |

> 🟡 Note: Problem 11 has an error in the provided answer. The correct answer is $ \frac{7}{2} $, not $ \frac{215}{2} $.