Solving logarithmic equations with examples and a question prompt.

Let's solve each of the logarithmic equations step by step. We'll go through them one at a time, using logarithmic properties and algebraic manipulation.

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1.

$$
\log_5(x + 6) = \log_5(2x + 4)
$$

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

$$
x + 6 = 2x + 4
$$

Solve for $ x $:

$$
6 - 4 = 2x - x \Rightarrow 2 = x
$$

Now check if this value is valid (i.e., arguments of logs are positive):

- $ x + 6 = 2 + 6 = 8 > 0 $
- $ 2x + 4 = 4 + 4 = 8 > 0 $

✔ Valid solution.

Answer: $ x = 2 $

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2.

$$
\log(3x - 2) = 2
$$

This is a common logarithm (base 10), so:

$$
\log_{10}(3x - 2) = 2
$$

Convert to exponential form:

$$
3x - 2 = 10^2 = 100
$$

$$
3x = 102 \Rightarrow x = 34
$$

Check domain: $ 3x - 2 = 3(34) - 2 = 102 - 2 = 100 > 0 $

✔ Valid.

Answer: $ x = 34 $

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3.

$$
\log_3(9x) - \log_3(x - 8) = 4
$$

Use log property: $ \log_b A - \log_b B = \log_b\left(\frac{A}{B}\right) $

$$
\log_3\left(\frac{9x}{x - 8}\right) = 4
$$

Convert to exponential form:

$$
\frac{9x}{x - 8} = 3^4 = 81
$$

Now solve:

$$
\frac{9x}{x - 8} = 81
$$

Multiply both sides by $ x - 8 $:

$$
9x = 81(x - 8) = 81x - 648
$$

$$
9x - 81x = -648 \Rightarrow -72x = -648 \Rightarrow x = 9
$$

Check domain:

- $ 9x = 81 > 0 $
- $ x - 8 = 1 > 0 $

✔ Valid.

Answer: $ x = 9 $

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4.

$$
\log_5(x - 1) + \log_5(x + 3) - 1 = 0
$$

Rewriting:

$$
\log_5(x - 1) + \log_5(x + 3) = 1
$$

Use log property: $ \log_b A + \log_b B = \log_b(AB) $

$$
\log_5[(x - 1)(x + 3)] = 1
$$

Convert to exponential:

$$
(x - 1)(x + 3) = 5^1 = 5
$$

Expand left side:

$$
x^2 + 3x - x - 3 = x^2 + 2x - 3 = 5
$$

$$
x^2 + 2x - 8 = 0
$$

Factor:

$$
(x + 4)(x - 2) = 0 \Rightarrow x = -4 \text{ or } x = 2
$$

Now check domain:

- $ x - 1 > 0 \Rightarrow x > 1 $
- $ x + 3 > 0 \Rightarrow x > -3 $

So $ x > 1 $. Thus, $ x = -4 $ is invalid.

Only $ x = 2 $ is valid.

Check:

- $ x - 1 = 1 > 0 $
- $ x + 3 = 5 > 0 $

✔ Valid.

Answer: $ x = 2 $

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5.

$$
\log_2(3 - x) + \log_2(5 + 2x) = \log_2 14
$$

Combine logs on left:

$$
\log_2[(3 - x)(5 + 2x)] = \log_2 14
$$

So arguments must be equal:

$$
(3 - x)(5 + 2x) = 14
$$

Expand:

$$
3(5 + 2x) - x(5 + 2x) = 15 + 6x - 5x - 2x^2 = 15 + x - 2x^2
$$

Set equal to 14:

$$
15 + x - 2x^2 = 14
\Rightarrow -2x^2 + x + 1 = 0
$$

Multiply both sides by -1:

$$
2x^2 - x - 1 = 0
$$

Factor:

$$
(2x + 1)(x - 1) = 0 \Rightarrow x = -\frac{1}{2} \text{ or } x = 1
$$

Now check domain:

- $ 3 - x > 0 \Rightarrow x < 3 $
- $ 5 + 2x > 0 \Rightarrow x > -\frac{5}{2} $

So $ x \in (-2.5, 3) $

Check both solutions:

1. $ x = -\frac{1}{2} $:
- $ 3 - (-0.5) = 3.5 > 0 $
- $ 5 + 2(-0.5) = 5 - 1 = 4 > 0 $
✔ Valid

2. $ x = 1 $:
- $ 3 - 1 = 2 > 0 $
- $ 5 + 2(1) = 7 > 0 $
✔ Valid

Now verify in original equation:

#### For $ x = -\frac{1}{2} $:

Left: $ \log_2(3 + 0.5) + \log_2(5 - 1) = \log_2(3.5) + \log_2(4) $

$ \log_2(3.5 \times 4) = \log_2(14) $ ✔

#### For $ x = 1 $:

$ \log_2(2) + \log_2(7) = 1 + \log_2(7) $

But $ \log_2(14) = \log_2(2 \cdot 7) = 1 + \log_2(7) $ ✔

Both satisfy!

So both solutions are valid.

Answers: $ x = -\frac{1}{2} $ or $ x = 1 $

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✔ Final Answers:

1. $ x = 2 $
2. $ x = 34 $
3. $ x = 9 $
4. $ x = 2 $
5. $ x = -\frac{1}{2} $ or $ x = 1 $

Let me know if you'd like these summarized!