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

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. $12\log(3x+4) - 4\log(2x+3) + \log(x-1) = 2$

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

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, $x = 9$.

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