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Trick Questions Based on Log Rules (page 4 of 5) Sections: Basic log rules, Expanding, Simplifying, Trick questions, ChangeofBase 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.
Since 10 = 2 × 5, then: log_{b}(10) = log_{b}(2 × 5) = log_{b}(2) + log_{b}(5) Since I have the values for log_{b}(2) and log_{b}(5), I can evaluate: log_{b}(2) + log_{b}(5) = 0.3869 + 0.8982 = 1.2851 Then log_{b}(10) = 1.2851.
Since 9 = 3^{2}, then: log_{b}(9) = log_{b}(3^{2}) = 2log_{b}(3) Since I have the value for log_{b}(3), then I can evaluate: 2log_{b}(3) = 2(0.6131) = 1.2262 Then log_{b}(9) = 1.2262.
This one is a bit more complicated, but, after fiddling with the numbers for a bit, I notice that 7.5 = 15 ÷ 2, so: log_{b}(7.5) = log_{b}(15 ÷ 2) = log_{b}(15) – log_{b}(2) And 15 = 5 × 3, so: Copyright © Elizabeth Stapel 20022011 All Rights Reserved log_{b}(15)
– log_{b}(2)
And now I can evaluate: log_{b}(5)
+ log_{b}(3) – log_{b}(2)
Then log_{b}(7.5) = 1.1244.
Since 6 = 2 × 3, then: log_{b}(6) = log_{b}(2 × 3) = log_{b}(2) + log_{b}(3) Since I have these values, I can evaluate: log_{b}(2) + log_{b}(3) = 0.3869 + 0.6131 = 1.0000 Then log_{b}(6) = 1.0000. Hmm... that was interesting. I got that log_{b}(6) = 1. Using The Relationship, I get: log_{b}(6)
= 1 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. << Previous Top  1  2  3  4  5  Return to Index Next >>



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