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Let's solve these logarithmic equations one by one.

### 1. \(\ln x = -3\)
To solve this, we rewrite the natural logarithm in exponential form:

\[
x = e^{-3}
\]

Thus, \(x = \frac{1}{e^3} \approx 0.0498\).

### 2. \(\log(3x - 2) = 2\)
Rewrite in exponential form using base 10:

\[
3x - 2 = 10^2 = 100
\]

Now, solve for \(x\):

\[
3x = 100 + 2 = 102 \quad \Rightarrow \quad x = \frac{102}{3} = 34
\]

### 3. \(2 \log x = \log 2 + \log(3x - 4)\)
Use the properties of logarithms to simplify:

\[
\log x^2 = \log(2(3x - 4))
\]

Since the logarithms are equal, we can set the arguments equal:

\[
x^2 = 2(3x - 4)
\]

Expanding and solving the quadratic equation:

\[
x^2 = 6x - 8 \quad \Rightarrow \quad x^2 - 6x + 8 = 0
\]

Using the quadratic formula:

\[
x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2}
\]

So, \(x = \frac{6 + 2}{2} = 4\) or \(x = \frac{6 - 2}{2} = 2\).

Thus, \(x = 4\) or \(x = 2\).

### 4. \(\log x + \log(x - 1) = \log(4x)\)
Combine the logarithms:

\[
\log(x(x - 1)) = \log(4x)
\]

Since the logarithms are equal, set the arguments equal:

\[
x(x - 1) = 4x
\]

Expand and simplify:

\[
x^2 - x = 4x \quad \Rightarrow \quad x^2 - 5x = 0 \quad \Rightarrow \quad x(x - 5) = 0
\]

Thus, \(x = 0\) or \(x = 5\). However, \(x = 0\) is not a valid solution for a logarithmic equation, so the only solution is \(x = 5\).

### 5. \(\log_3(x + 25) - \log_3(x - 1) = 3\)
Use the properties of logarithms:

\[
\log_3\left(\frac{x + 25}{x - 1}\right) = 3
\]

Rewrite in exponential form:

\[
\frac{x + 25}{x - 1} = 3^3 = 27
\]

Now solve for \(x\):

\[
x + 25 = 27(x - 1) \quad \Rightarrow \quad x + 25 = 27x - 27 \quad \Rightarrow \quad 25 + 27 = 26x \quad \Rightarrow \quad x = \frac{52}{26} = 2
\]

### 6. \(\log_9(x - 5) + \log_9(x + 3) = 1\)
Combine the logarithms:

\[
\log_9((x - 5)(x + 3)) = 1
\]

Rewrite in exponential form:

\[
(x - 5)(x + 3) = 9^1 = 9
\]

Expand the quadratic:

\[
x^2 - 5x + 3x - 15 = 9 \quad \Rightarrow \quad x^2 - 2x - 15 = 9 \quad \Rightarrow \quad x^2 - 2x - 24 = 0
\]

Solve using the quadratic formula:

\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-24)}}{2(1)} = \frac{2 \pm \sqrt{4 + 96}}{2} = \frac{2 \pm \sqrt{100}}{2} = \frac{2 \pm 10}{2}
\]

Thus, \(x = \frac{2 + 10}{2}

Solve the following logarithmic equations: 1. ln x = -3, 2. log(3x - 2) = 2, 3. 2 log x = log 2 + log(3x - 4), 4. log x + log(x - 1) = log(4x), 5. log3(x + 25) - log3(x - 1) = 3, 6. log9(x - 5) + log9(x + 3) = 1, 7. log x + log (x - 3) = 1, 8. log2(x - 2) + log2(x + 1) = 2.

Step-by-step solutions for solving logarithmic equations, covering key concepts such as properties of logarithms, exponential functions, and quadratic equations. Learn how to solve equations like ln(x) = -3, log(3x - 2) = 2, and more.

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