Math Problem Statement

If log10 ab and log10 b = c and log10 100 = c, then which of the following might be true?

452 +852 > a+b

5ck = a100 for some integer value of k

(452)(852) > ab

None of the above

Solution

Given the equations:

log10(ab)=log10(b)=candlog10(100)=c\log_{10}(ab) = \log_{10}(b) = c \quad \text{and} \quad \log_{10}(100) = c

Let's analyze them step by step:

  1. Interpreting the equations:

    • log10(100)=c\log_{10}(100) = c implies c=2c = 2, because 102=10010^2 = 100.
    • log10(ab)=log10(b)=c\log_{10}(ab) = \log_{10}(b) = c implies: log10(ab)=c=2andlog10(b)=2\log_{10}(ab) = c = 2 \quad \text{and} \quad \log_{10}(b) = 2
    • If log10(b)=2\log_{10}(b) = 2, then b=100b = 100.
    • If log10(ab)=2\log_{10}(ab) = 2, then ab=102=100ab = 10^2 = 100.
  2. From ab=100ab = 100 and b=100b = 100:

    a×100=100    a=1a \times 100 = 100 \implies a = 1

Thus, a=1a = 1 and b=100b = 100.

  1. Evaluating the options:

    • Option 1: 452+852>a+b452 + 852 > a + b

      Substituting a=1a = 1 and b=100b = 100: 452+852=1304anda+b=1+100=101452 + 852 = 1304 \quad \text{and} \quad a + b = 1 + 100 = 101 Clearly, 1304>1011304 > 101. So, this statement might be true.

    • Option 2: 5ck=a1005ck = a^{100} for some integer value of kk

      Substituting c=2c = 2 and a=1a = 1: 5×2×k=1100=15 \times 2 \times k = 1^{100} = 1 However, 5×2×k=10k5 \times 2 \times k = 10k. For kk to be an integer, 10k=110k = 1 is not possible. So, this statement cannot be true.

    • Option 3: (452)(852)>ab(452)(852) > ab

      Substituting ab=100ab = 100: (452)(852)=385104andab=100(452)(852) = 385104 \quad \text{and} \quad ab = 100 Clearly, 385104>100385104 > 100. So, this statement might be true.

    • Option 4: None of the above

      Since two of the statements might be true, this option is incorrect.

Conclusion:

Both Option 1 and Option 3 might be true.

Would you like more details or have any questions?

Here are 5 related questions to explore:

  1. What are the properties of logarithms?
  2. How can logarithmic equations be used to solve exponential equations?
  3. What are the key differences between natural logarithms and common logarithms?
  4. How would the solution change if the base of the logarithm was different?
  5. How can logarithms be applied in solving real-world problems?

Tip: Remember that logarithms are the inverses of exponentiation. This relationship is crucial when solving logarithmic equations.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponential Equations

Formulas

Logarithm properties
Exponential equations

Theorems

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Suitable Grade Level

High School