Algebra factoring practice worksheet with equations and solutions.

It looks like you've uploaded an image containing a math problem, but the text is not clearly readable due to formatting or resolution issues. However, I can see some mathematical expressions and equations, such as:

- $ 2 \cdot 16^x $
- $ 4^{x+1} = 16^{x-1} $
- $ 8^{x-1} = 27 $
- $ 3^{2x} = 9^{x-1} $

From these, it seems like the task involves solving exponential equations using properties of exponents.

Let me solve one of the most common types of problems that appear in such images — for example, solving:

Example Problem:

Solve:
$$
4^{x+1} = 16^{x-1}
$$

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Step-by-step Solution:

#### Step 1: Express both sides with the same base.
We know:
- $ 4 = 2^2 $
- $ 16 = 2^4 $

So rewrite both sides in terms of base 2:

$$
(2^2)^{x+1} = (2^4)^{x-1}
$$

#### Step 2: Simplify exponents using power rule $ (a^m)^n = a^{mn} $:

$$
2^{2(x+1)} = 2^{4(x-1)}
$$

$$
2^{2x + 2} = 2^{4x - 4}
$$

#### Step 3: Since the bases are equal, set the exponents equal:

$$
2x + 2 = 4x - 4
$$

#### Step 4: Solve for $ x $:

$$
2 + 4 = 4x - 2x
$$
$$
6 = 2x
$$
$$
x = 3
$$

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Final Answer:

$$
\boxed{x = 3}
$$

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If your image contains other similar equations (like $ 8^{x-1} = 27 $), here's how to solve that:

Solve: $ 8^{x-1} = 27 $

Note:
- $ 8 = 2^3 $
- $ 27 = 3^3 $

But different bases, so take logarithms:

$$
\log(8^{x-1}) = \log(27)
$$
$$
(x-1)\log(8) = \log(27)
$$
$$
x - 1 = \frac{\log(27)}{\log(8)}
$$
$$
x = 1 + \frac{\log(27)}{\log(8)} = 1 + \frac{\log(3^3)}{\log(2^3)} = 1 + \frac{3\log 3}{3\log 2} = 1 + \frac{\log 3}{\log 2}
$$

This is approximately:
$$
x \approx 1 + \frac{0.477}{0.301} \approx 1 + 1.585 = 2.585
$$

But if the equation was meant to be $ 8^{x-1} = 2^{x+1} $, then we could solve it exactly.

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Please clarify or confirm which specific problem from the image you'd like solved (e.g., "solve $ 4^{x+1} = 16^{x-1} $" or another). If you can describe or type out the exact problem, I’ll give you a precise solution with full explanation.