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These are exponential inequalities. I'll walk through each one step-by-step to solve them.

1. **\( 5^{x+2} > 25 \)**  
   Rewrite 25 as \(5^2\):  
   \( 5^{x+2} > 5^2 \)  
   Since the bases are the same, compare the exponents:  
   \( x + 2 > 2 \)  
   \( x > 0 \)

2. **\( 9^x < 81 \)**  
   Rewrite 81 as \(9^2\):  
   \( 9^x < 9^2 \)  
   \( x < 2 \)

3. **\( 2^{x-1} < 4^3 \)**  
   Rewrite \(4\) as \(2^2\), so \(4^3 = (2^2)^3 = 2^6\):  
   \( 2^{x-1} < 2^6 \)  
   \( x - 1 < 6 \)  
   \( x < 7 \)

4. **\( 7^{4x+3} \geq 7^{x-6} \)**  
   Since the bases are the same, compare the exponents:  
   \( 4x + 3 \geq x - 6 \)  
   \( 3x \geq -9 \)  
   \( x \geq -3 \)

5. **\( 3^{6x+2} \geq 9^{2x} \)**  
   Rewrite \(9\) as \(3^2\), so \(9^{2x} = (3^2)^{2x} = 3^{4x}\):  
   \( 3^{6x+2} \geq 3^{4x} \)  
   \( 6x + 2 \geq 4x \)  
   \( 2x \geq -2 \)  
   \( x \geq -1 \)

6. **\( \left(\frac{1}{2}\right)^{3x+4} < \left(\frac{1}{4}\right)^{2x-1} \)**  
   Rewrite \(\frac{1}{4}\) as \((\frac{1}{2})^2\):  
   \( \left(\frac{1}{2}\right)^{3x+4} < \left(\frac{1}{2}\right)^{2(2x-1)} \)  
   \( \left(\frac{1}{2}\right)^{3x+4} < \left(\frac{1}{2}\right)^{4x-2} \)  
   Since the bases are the same and are fractions, the inequality sign flips when comparing exponents:  
   \( 3x + 4 > 4x - 2 \)  
   \( 6 > x \)  
   \( x < 6 \)

7. **\( \left(\frac{1}{9}\right)^{2x-1} \geq \left(\frac{1}{27}\right)^x \)**  
   Rewrite \(\frac{1}{9}\) as \((\frac{1}{3})^2\) and \(\frac{1}{27}\) as \((\frac{1}{3})^3\):  
   \( \left(\frac{1}{3}\right)^{2(2x-1)} \geq \left(\frac{1}{3}\right)^{3x} \)  
   \( \left(\frac{1}{3}\right)^{4x-2} \geq \left(\frac{1}{3}\right)^{3x} \)  
   Since the bases are the same, compare the exponents:  
   \( 4x - 2 \geq 3x \)  
   \( x \geq 2 \)

8. **\( \left(\frac{1}{8}\right)^{2x+1} \geq \left(\frac{1}{4}\right)^{x+2} \)**  
   Rewrite \(\frac{1}{8}\) as \((\frac{1}{2})^3\) and \(\frac{1}{4}\) as \((\frac{1}{2})^2\):  
   \( \left(\frac{1}{2}\right)^{3(2x+1)} \geq \left(\frac{1}{2}\right)^{2(x+2)} \)  
   \( \left(\frac{1}{2}\right)^{6x+3} \geq \left(\frac{1}{2}\right)^{2x+4} \)  
   Compare the exponents:  
   \( 6x + 3 \leq 2x + 4 \)  
   \( 4x \leq 1 \)  
   \( x \leq \frac{1}{4} \)

Let me know if you'd like further details or have any questions on any step!

Here are five related questions to expand on this topic:

1. How would you solve exponential inequalities with different bases that cannot be expressed as powers of each other?
2. What happens to the inequality sign when dealing with bases between 0 and 1?
3. How do logarithmic functions help solve more complex exponential inequalities?
4. Can you solve inequalities involving irrational bases, such as \( e^x \)?
5. What are some real-world applications of exponential inequalities?

**Tip:** Remember, when you have an inequality with a fractional base, reversing the inequality sign is necessary when comparing exponents, as fractional powers have an inverse relationship with magnitude.

Solve the exponential inequalities in the given image.

Learn how to solve exponential inequalities, such as 5^(x+2) > 25, 9^x < 81, and similar problems. This guide includes step-by-step solutions, key concepts in exponential inequalities, and essential formulas for comparing exponents with the same or fractional bases. Suitable for high school students in Grades 9-12.

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