Math Problem Statement

evaluate the following logarithmic functions

Evaluate the following logarithmic functions.

Solution

Let's evaluate each logarithmic function step-by-step:

  1. f(x)=log24f(x) = \log_2 4

    • We want xx such that 2x=42^x = 4.
    • Since 4=224 = 2^2, x=2x = 2.
    • Answer: f(x)=2f(x) = 2.
  2. f(x)=log255f(x) = \log_{25} 5

    • We want xx such that 25x=525^x = 5.
    • Since 25=5225 = 5^2, 5=251/25 = 25^{1/2}.
    • So, x=12x = \frac{1}{2}.
    • Answer: f(x)=12f(x) = \frac{1}{2}.
  3. f(x)=log6464f(x) = \log_{64} 64

    • We want xx such that 64x=6464^x = 64.
    • x=1x = 1 since 641=6464^1 = 64.
    • Answer: f(x)=1f(x) = 1.
  4. f(x)=log1/319f(x) = \log_{1/3} \frac{1}{9}

    • We want xx such that (13)x=19\left(\frac{1}{3}\right)^x = \frac{1}{9}.
    • Since 19=(13)2\frac{1}{9} = \left(\frac{1}{3}\right)^2, x=2x = 2.
    • Answer: f(x)=2f(x) = 2.
  5. f(x)=log1/31f(x) = \log_{1/3} 1

    • We want xx such that (13)x=1\left(\frac{1}{3}\right)^x = 1.
    • Any number to the power of zero is 1, so x=0x = 0.
    • Answer: f(x)=0f(x) = 0.
  6. f(x)=log1/327f(x) = \log_{1/3} 27

    • We want xx such that (13)x=27\left(\frac{1}{3}\right)^x = 27.
    • 27=3327 = 3^3, so 13\frac{1}{3} needs to be raised to 3-3 to equal 27.
    • x=3x = -3.
    • Answer: f(x)=3f(x) = -3.
  7. f(x)=log164f(x) = \log_{16} 4

    • We want xx such that 16x=416^x = 4.
    • Since 16=4216 = 4^2, we can express 16x=(42)x=42x16^x = (4^2)^x = 4^{2x}.
    • Thus, 42x=44^{2x} = 4 implies 2x=12x = 1, so x=12x = \frac{1}{2}.
    • Answer: f(x)=12f(x) = \frac{1}{2}.
  8. f(x)=log1/313f(x) = \log_{1/3} \frac{1}{3}

    • We want xx such that (13)x=13\left(\frac{1}{3}\right)^x = \frac{1}{3}.
    • x=1x = 1 since (13)1=13\left(\frac{1}{3}\right)^1 = \frac{1}{3}.
    • Answer: f(x)=1f(x) = 1.
  9. f(x)=log218f(x) = \log_2 \frac{1}{8}

    • We want xx such that 2x=182^x = \frac{1}{8}.
    • Since 18=23\frac{1}{8} = 2^{-3}, x=3x = -3.
    • Answer: f(x)=3f(x) = -3.
  10. f(x)=log1010f(x) = \log_{10} 10

    • We want xx such that 10x=1010^x = 10.
    • x=1x = 1 since 101=1010^1 = 10.
    • Answer: f(x)=1f(x) = 1.

Would you like further details on any of these solutions, or do you have any questions?

Here are some related questions:

  1. How would you solve logarithmic equations involving irrational bases?
  2. Can you explain the change of base formula and its applications?
  3. How does the concept of logarithms extend to complex numbers?
  4. What are some real-world applications of logarithmic functions?
  5. How do logarithmic functions relate to exponential growth and decay?

Tip: When dealing with fractional bases, remember that negative exponents invert the base.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponents
Properties of Logarithms

Formulas

Change of Base Formula
Logarithmic identity: log_b(b^k) = k
Negative exponents and fractional bases

Theorems

Logarithmic and Exponential Relationship

Suitable Grade Level

High School