Questions: Solve the following absolute value inequality 32x-3+1 >= 10 0 x>0 and x>3 (-∞, 0] U [3, ∞) no solution (-∞,-3] ∪ [0, ∞) [0,3] Solve: 9^(2x-1)=27 -3 / 4 O 3 / 4 .5 / 4 no solution

Solve the following absolute value inequality 32x-3+1 >= 10 0 x>0 and x>3 (-∞, 0] U [3, ∞) no solution (-∞,-3] ∪ [0, ∞) [0,3] Solve: 9^(2x-1)=27 -3 / 4 O 3 / 4 .5 / 4 no solution
Transcript text: Solve the following absolute value inequality \[ 3|2 x-3|+1 \geq 10 \] $0 x>0$ and $x>3$ $(-\infty, 0]$ U $[3, \infty)$ no solution $(-\infty,-3] \cup[0, \infty)$ [0,3] Solve: $9^{(2 x-1)}=27$ $-3 / 4$ O $3 / 4$ $.5 / 4$ no solution
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Solution

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Solution Steps

Question 2
Step 1: Simplify the Inequality

The given inequality is: \[ 3|2x - 3| + 1 \geq 10 \]

First, subtract 1 from both sides: \[ 3|2x - 3| \geq 9 \]

Step 2: Divide by 3

Divide both sides by 3: \[ |2x - 3| \geq 3 \]

Step 3: Solve the Absolute Value Inequality

The inequality \(|2x - 3| \geq 3\) can be split into two separate inequalities:

  1. \(2x - 3 \geq 3\)
  2. \(2x - 3 \leq -3\)
Solving \(2x - 3 \geq 3\):

Add 3 to both sides: \[ 2x \geq 6 \]

Divide by 2: \[ x \geq 3 \]

Solving \(2x - 3 \leq -3\):

Add 3 to both sides: \[ 2x \leq 0 \]

Divide by 2: \[ x \leq 0 \]

Step 4: Combine the Solutions

The solution to the inequality is: \[ x \leq 0 \quad \text{or} \quad x \geq 3 \]

In interval notation, this is: \[ (-\infty, 0] \cup [3, \infty) \]

Final Answer for Question 2

\[ \boxed{(-\infty, 0] \cup [3, \infty)} \]

Question 3
Step 1: Express Both Sides with the Same Base

The given equation is: \[ 9^{(2x-1)} = 27 \]

Express 9 and 27 as powers of 3: \[ (3^2)^{(2x-1)} = 3^3 \]

Step 2: Simplify the Exponents

Simplify the left side: \[ 3^{2(2x-1)} = 3^3 \]

This implies: \[ 2(2x-1) = 3 \]

Step 3: Solve for \(x\)

Distribute the 2: \[ 4x - 2 = 3 \]

Add 2 to both sides: \[ 4x = 5 \]

Divide by 4: \[ x = \frac{5}{4} \]

Final Answer for Question 3

\[ \boxed{\frac{5}{4}} \]

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