The first two properties listed here can be a little

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Unformatted text preview: As always let’s first convert to exponential form. © 2007 Paul Dawkins 286 http://tutorial.math.lamar.edu/terms.aspx College Algebra log 5 1 =? 125 Þ 5? = 1 125 First, notice that the only way that we can raise an integer to an integer power and get a fraction as an answer is for the exponent to be negative. So, we know that the exponent has to be negative. Now, let’s ignore the fraction for a second and ask 5? = 125 . In this case if we cube 5 we will get 125. So, it looks like we have the following, 1 = -3 125 log 5 because 5- 3 = 1 1 = 3 5 125 [Return to Problems] (e) log 1 81 3 Converting this logarithm to exponential form gives, ? log 1 81 = ? æ1ö ç ÷ = 81 è3ø Þ 3 Now, just like the previous part, the only way that this is going to work out is if the exponent is negative. Then all we need to do is recognize that 34 = 81 and we can see that, -4 log 1 81 = -4 because 3 4 æ1ö æ3ö 4 ç ÷ = ç ÷ = 3 = 81 è 3ø è1ø [Return to Problems] (f) log 3 2 27 8 Here is the answer to this one. 27 log 3 =3 28 3 becau...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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