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I see that you uploaded a photo of mathematical problems. Let me transcribe the questions for clarity and solve them for you.

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### Problems:
1. \( \log(x^2 - 1) - \log(x - 1) = 1 \)
2. \( 2\log(x+6) - \log(2x - 1) = \log(x + 1) \)
3. \( 2^x = 128 \)
4. \( 3x + 12 = 6 \)
5. \( x^2 + 7x = -15 \)
6. \( x^2 - 5x - 54 = 0 \)
7. \( 3x^2 - 14x + 5 = 0 \)
8. \( x^2 - 10x + 24 = 0 \)
9. \( 12\log(3x+4) - 4\log(2x+3) + \log(x-1) = 2 \)

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I'll solve these equations step by step, starting with the first one.

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### 1. \( \log(x^2 - 1) - \log(x - 1) = 1 \)

Using the logarithmic identity \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \), we get:

\[
\log\left(\frac{x^2 - 1}{x - 1}\right) = 1
\]

Since \( x^2 - 1 = (x - 1)(x + 1) \), this becomes:

\[
\log(x + 1) = 1
\]

Now, converting the logarithm to exponential form:

\[
x + 1 = 10^1 = 10
\]

Thus, \( x = 9 \).

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I'll solve the remaining equations next. Do you want more detailed steps for this problem or the rest of them?

Solve the following equations, giving your answers correct to 3 decimal places where necessary:
1. log(x^2 - 1) - log(x - 1) = 1
2. 2log(x+6) - log(2x - 1) = log(x + 1)
3. 2^x = 128
4. 3x + 12 = 6
5. x^2 + 7x = -15
6. x^2 - 5x - 54 = 0
7. 3x^2 - 14x + 5 = 0
8. x^2 - 10x + 24 = 0
9. 12log(3x+4) - 4log(2x+3) + log(x-1) = 2

Learn how to solve various types of equations including logarithmic, exponential, and quadratic equations with detailed step-by-step solutions. This guide covers key mathematical concepts such as properties of logarithms, exponential equations, and the quadratic formula theorem. Suitable for students in Grades 9-12.

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