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Trick Questions Based on Log Rules (page 4 of 5)

Sections: Basic log rules, Expanding, Simplifying, Trick questions, Change-of-Base formula


You should expect to need to know these log rules, because there is a certain type of question that the teacher can put on the test to make sure you know how to use the rules. Warning: You won't be able to "cheat" with your calculator.

  • Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. Using these values, evaluate logb(10).

    Since 10 = 2 5, then:

      logb(10) = logb(2 5) = logb(2) + logb(5)

    Since I have the values for logb(2) and logb(5), I can evaluate:

      logb(2) + logb(5) = 0.3869 + 0.8982 = 1.2851

    Then logb(10) = 1.2851.

  • Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. Using these values, evaluate logb(9).

    Since 9 = 32, then:

      logb(9) = logb(32) = 2logb(3)

    Since I have the value for logb(3), then I can evaluate:

      2logb(3) = 2(0.6131) = 1.2262

    Then logb(9) = 1.2262.

  • Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. Using these values, evaluate logb(7.5).

    This one is a bit more complicated, but, after fiddling with the numbers for a bit, I notice that 7.5 = 15  2, so:

      logb(7.5) = logb(15 2) = logb(15) logb(2)

    And 15 = 5 3, so:   Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

      logb(15) logb(2)
           = [logb(5) + logb(3)] logb(2)
           = logb(5) + logb(3) logb(2)

    And now I can evaluate:

      logb(5) + logb(3) logb(2)
           = 0.8982 + 0.6131 0.3869
           = 1.1244

    Then logb(7.5) = 1.1244.

  • Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. Using these values, evaluate logb(6).

    Since 6 = 2 3, then:

      logb(6) = logb(2 3) = logb(2) + logb(3)

    Since I have these values, I can evaluate:

      logb(2) + logb(3) = 0.3869 + 0.6131 = 1.0000

    Then logb(6) = 1.0000.

Hmm... that was interesting. I got that logb(6) = 1. Using The Relationship, I get:

    logb(6) = 1
    b1 = 6
    b = 6

So now I know that their mysterious unnamed base "b" was actually 6! But they will not usually give you problems that let you figure out the base like this.

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Cite this article as:

Stapel, Elizabeth. "Trick Questions Based on Log Rules." Purplemath. Available from
    http://www.purplemath.com/modules/logrules4.htm. Accessed
 

 



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