Math Problem Statement

. 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(4x+6) = 2 d. logx(16) = −2

c. log(3x−2) = 1 d. logx(16) = −2

Solution

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

a. log8(1162)=x\log_8\left(\frac{1}{16\sqrt{2}}\right) = x

  1. Restrictions on xx:

    • Since xx 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: 8x=11628^x = \frac{1}{16\sqrt{2}}

  3. Solve for xx: First, express everything in terms of powers of 2: 8=238 = 2^3 2=21/2\sqrt{2} = 2^{1/2} 16=2416 = 2^4 Therefore, 8x=(23)x=23x8^x = (2^3)^x = 2^{3x} 1162=12421/2=124.5=24.5\frac{1}{16\sqrt{2}} = \frac{1}{2^4 \cdot 2^{1/2}} = \frac{1}{2^{4.5}} = 2^{-4.5}

    Equate the exponents: 23x=24.52^{3x} = 2^{-4.5} 3x=4.53x = -4.5 x=1.5x = -1.5

b. log3(4x+6)=2\log_3(4x + 6) = 2

  1. Restrictions on xx:

    • The argument of the logarithm must be positive: 4x+6>04x + 6 > 0
    • Therefore, x>64=1.5x > -\frac{6}{4} = -1.5
  2. Convert to exponential form: 32=4x+63^2 = 4x + 6

  3. Solve for xx: 9=4x+69 = 4x + 6 4x=964x = 9 - 6 4x=34x = 3 x=34x = \frac{3}{4}

c. log(3x2)=1\log(3x - 2) = 1

  1. Restrictions on xx:

    • The argument of the logarithm must be positive: 3x2>03x - 2 > 0
    • Therefore, x>23x > \frac{2}{3}
  2. Convert to exponential form: 101=3x210^1 = 3x - 2

  3. Solve for xx: 10=3x210 = 3x - 2 3x=10+23x = 10 + 2 3x=123x = 12 x=4x = 4

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

  1. Restrictions on xx:

    • The base of the logarithm must be positive and not equal to 1: x>0x > 0 and x1x \neq 1
    • 16 must be expressible as a power of xx.
  2. Convert to exponential form: x2=16x^{-2} = 16

  3. Solve for xx: 1x2=16\frac{1}{x^2} = 16 x2=116x^2 = \frac{1}{16} x=±14x = \pm \frac{1}{4}

    Considering the restriction x>0x > 0: x=14x = \frac{1}{4}

Summary:

  • a. x=1.5x = -1.5
  • b. x=34x = \frac{3}{4}
  • c. x=4x = 4
  • d. x=14x = \frac{1}{4}

Would you like more details or have any questions?


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.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponential equations

Formulas

Logarithm conversion to exponential form

Theorems

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Suitable Grade Level

Grades 10-12