In these cases it is almost always best to deal with

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Unformatted text preview: f) log 8 1 Solution To do the first four evaluations we just need to remember what the notation for these are and what base is implied by the notation. The final two evaluations are to illustrate some of the properties of all logarithms that we’ll be looking at eventually. (a) log1000 = 3 because 103 = 1000 . (b) log (c) ln 1 1 1 . = -2 because 10 -2 = 2 = 100 10 100 1 1 = -1 because e -1 = . e e 1 1 (d) ln e = because e 2 = e . Notice that with this one we are really just acknowledging a 2 change of notation from fractional exponent into radical form. (e) log 34 34 = 1 because 341 = 34 . Notice that this one will work regardless of the base that we’re using. (f) log 8 1 = 0 because 80 = 1 . Again, note that the base that we’re using here won’t change the answer. So, when evaluating logarithms all that we’re really asking is what exponent did we put onto the base to get the number in the logarithm. Now, before we get into some of the properties of logarithms let’s first do a coupl...
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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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