https://file.aiquickdraw.com/mathbot/file/1727004708278t4lsy1ka.jpg

I can help break down the questions from the image and explain the solutions step by step.

### 1. First set of questions:
#### 1a. \( (a + 4b)^2 \)
This is a simple binomial expansion using the identity:
\[
(a + 4b)^2 = a^2 + 2ab \cdot 4 + (4b)^2
\]
Thus, the expanded form is:
\[
a^2 + 8ab + 16b^2
\]
The answer provided in the image appears to be "8", but that's unrelated to the expression itself, so it's possibly just part of feedback or grading.

#### 1b. Simplify:
\[
\frac{\sqrt{50} + \sqrt{18}}{\sqrt{32 \times 2}}
\]
First, simplify the square roots:
\[
\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}, \quad \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}, \quad \sqrt{32 \times 2} = \sqrt{64} = 8
\]
Now, substitute into the expression:
\[
\frac{5\sqrt{2} + 3\sqrt{2}}{8} = \frac{8\sqrt{2}}{8} = \sqrt{2}
\]
Thus, the simplified result is \( \sqrt{2} \), and the answer provided is "5", possibly due to some grading or feedback.

### 2. Solve for \(x\) in the equation:
\[
25^x + 25^x + 25^x + 25^x + 25^x = 5^{2021}
\]
The left side can be simplified as:
\[
5 \times 25^x = 5 \times (5^2)^x = 5 \times 5^{2x} = 5^{1 + 2x}
\]
So, we now have:
\[
5^{1 + 2x} = 5^{2021}
\]
Equating the exponents:
\[
1 + 2x = 2021
\]
Solving for \(x\):
\[
2x = 2020 \quad \Rightarrow \quad x = 1010
\]
The answer provided in the image is "25", but the correct solution for \(x\) is 1010.

### 3. Simplify the expression:
\[
\frac{(3x^2y^{-2}z^3)^3}{(x^{-2}y^2z^{-2})^2}
\]
First, simplify each part:
- For the numerator:
\[
(3x^2y^{-2}z^3)^3 = 3^3x^{6}y^{-6}z^9 = 27x^6y^{-6}z^9
\]
- For the denominator:
\[
(x^{-2}y^2z^{-2})^2 = x^{-4}y^4z^{-4}
\]
Now, divide the terms:
\[
\frac{27x^6y^{-6}z^9}{x^{-4}y^4z^{-4}} = 27x^{6 - (-4)}y^{-6 - 4}z^{9 - (-4)} = 27x^{10}y^{-10}z^{13}
\]
Thus, the simplified expression is:
\[
27x^{10}y^{-10}z^{13}
\]
The answer provided is "10", but the simplified form is \(27x^{10}y^{-10}z^{13}\).

### 4. Simplify the logarithmic expression:
\[
2 \log 8 + 3 \log 9 + 5 \log 125
\]
We use the logarithmic identity: \( a \log b = \log b^a \), so this becomes:
\[
\log 8^2 + \log 9^3 + \log 125^5
\]
Simplify each term:
\[
8^2 = 64, \quad 9^3 = 729, \quad 125^5 = 3125^2 = 9765625
\]
Now, use the property \( \log a + \log b = \log (a \times b) \):
\[
\log(64 \times 729 \times 9765625)
\]
This results in a very large number, but the expression can be simplified further through calculation.

---

Would you like more details or clarifications on any of these solutions? Here are 5 related questions you might find interesting:

1. How would you simplify other binomial expansions, such as \( (a + b)^3 \)?
2. Can you solve logarithmic equations where the base is not 10, for example \( \log_2 8 \)?
3. What is the general rule for solving exponential equations like \( a^x = b \)?
4. How do you apply logarithmic properties to simplify complex expressions?
5. Can you find the derivative of the expression \( 3x^2y^{-2}z^3 \)?

**Tip:** Always break down complex expressions into simpler parts, applying rules such as factoring or logarithmic identities step by step.

Simplify various algebraic and logarithmic expressions, and solve for x in an equation involving exponents. The user provided four mathematical problems: binomial expansion, square root simplifications, solving an exponential equation, and simplifying a complex fraction and logarithmic expression.

Explore step-by-step solutions for simplifying exponential and logarithmic expressions. This guide covers binomial expansions, square root simplifications, and solving complex exponential and logarithmic equations. Suitable for high school students in Grades 10-12.

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