**.** Identify the restrictions on *x*. Convert each equation to the equivalent exponential form and

solve for *x*.

a. log8(1/16 sqaure root of 2)=x    b. log3(4*x*+6) = 2 d. log*x*(16) = −2

c. log(3*x*−2) = 1 d. log*x*(16) = −2

Let's go through each equation step-by-step, identifying the restrictions on \( x \), converting each to its exponential form, and solving for \( x \).

### a. \( \log_8\left(\frac{1}{16\sqrt{2}}\right) = x \)

1. **Restrictions on \( x \)**:
   - Since \( x \) appears on one side and doesn't directly influence the argument of the log function, there are no direct restrictions here.

2. **Convert to exponential form**:
   \[ 8^x = \frac{1}{16\sqrt{2}} \]

3. **Solve for \( x \)**:
   First, express everything in terms of powers of 2:
   \[ 8 = 2^3 \]
   \[ \sqrt{2} = 2^{1/2} \]
   \[ 16 = 2^4 \]
   Therefore,
   \[ 8^x = (2^3)^x = 2^{3x} \]
   \[ \frac{1}{16\sqrt{2}} = \frac{1}{2^4 \cdot 2^{1/2}} = \frac{1}{2^{4.5}} = 2^{-4.5} \]

   Equate the exponents:
   \[ 2^{3x} = 2^{-4.5} \]
   \[ 3x = -4.5 \]
   \[ x = -1.5 \]

### b. \( \log_3(4x + 6) = 2 \)

1. **Restrictions on \( x \)**:
   - The argument of the logarithm must be positive: \( 4x + 6 > 0 \)
   - Therefore, \( x > -\frac{6}{4} = -1.5 \)

2. **Convert to exponential form**:
   \[ 3^2 = 4x + 6 \]

3. **Solve for \( x \)**:
   \[ 9 = 4x + 6 \]
   \[ 4x = 9 - 6 \]
   \[ 4x = 3 \]
   \[ x = \frac{3}{4} \]

### c. \( \log(3x - 2) = 1 \)

1. **Restrictions on \( x \)**:
   - The argument of the logarithm must be positive: \( 3x - 2 > 0 \)
   - Therefore, \( x > \frac{2}{3} \)

2. **Convert to exponential form**:
   \[ 10^1 = 3x - 2 \]

3. **Solve for \( x \)**:
   \[ 10 = 3x - 2 \]
   \[ 3x = 10 + 2 \]
   \[ 3x = 12 \]
   \[ x = 4 \]

### d. \( \log_x(16) = -2 \)

1. **Restrictions on \( x \)**:
   - The base of the logarithm must be positive and not equal to 1: \( x > 0 \) and \( x \neq 1 \)
   - 16 must be expressible as a power of \( x \).

2. **Convert to exponential form**:
   \[ x^{-2} = 16 \]

3. **Solve for \( x \)**:
   \[ \frac{1}{x^2} = 16 \]
   \[ x^2 = \frac{1}{16} \]
   \[ x = \pm \frac{1}{4} \]

   Considering the restriction \( x > 0 \):
   \[ x = \frac{1}{4} \]

### Summary:

- **a.** \( x = -1.5 \)
- **b.** \( x = \frac{3}{4} \)
- **c.** \( x = 4 \)
- **d.** \( x = \frac{1}{4} \)

Would you like more details or have any questions?

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**8 Relative Questions:**

1. What are the general restrictions on the arguments of logarithms?
2. How do you convert a logarithmic equation to its exponential form?
3. Why is it necessary to express all terms with the same base when solving logarithmic equations?
4. Can logarithms have negative bases? Why or why not?
5. What is the significance of ensuring the argument of a logarithm is positive?
6. How do you handle equations where the logarithm has a variable base?
7. Why must the base of a logarithm be greater than zero and not equal to one?
8. What are some real-world applications of logarithmic equations?

**Tip:** When dealing with logarithmic equations, always check for potential extraneous solutions by ensuring the solutions satisfy the original restrictions on the domain.

Learn how to solve logarithmic equations step-by-step with detailed examples. Covering concepts like converting logarithms to exponential form and applying restrictions on logarithmic equations. Suitable for Grades 10-12 students.

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