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Sections 5.3-5.5 Logarithmic Functions (5.3) What is a logarithm??? LOGS ARE POWERS!!!! A logarithm or “log” of a number of a certain base is the exponent to which the base of the log must be raised in order to produce the number. The base cannot equal 1 and must be greater than 0. For instance, if logb(x) = c and b≠1 and b>0, then c is the specific exponent to which you must raise b in order to get x: bc = x Logarithmic Functions Why do we need logs? Let’s explore… 32 = 9 and 33 = 27 but what would we need to raise 3 to in order to get 20?? 3a = 20 that’s what logs tell us!! a = log320 Which two integers is log320 between? 2 and 3 Logarithmic Functions From the definition, we have stated that if logb(x) = c, then bc = x under the conditions that b≠1 and b>0. Why do we need to place any restrictions on b or x so this can make sense? Let’s try some values… logb(x) = c, so c b = x log2(8) = c so 2c = 8 c = 3, so far we are ok log1(5) = c so 1c = 5 Does not exist; 1c always equals 1 log-2(8) = c so (-2)c = 8 Does not exist; if c = 3, then (-2)3 = -8 log3(-9) = c so 3c = -9 Does not exist; 3c cannot be negative log2(0) = c so 2c = 0 Does not exist; 2c cannot equal 0 Summary: b≠1, b>0 and x>0 Log Properties (1) logbb = 1 (2) logb1 = 0 common log has base 10: log(x) = log10(x) natural log has base e: ln(x) = loge(x) Therefore… log10 = 1 lne = 1 Practice Evaluate, if possible. If not, state so. 1)log 5 (125) 2)log 32 (2) 3)log 4 (-16) 4)log 1 (27) 3 y = logb(x - h) + k When graphing logs we first need to identify and graph the asymptote. Earlier we discovered that the argument inside the log must be greater than 0. Therefore, x > h so the domain is (h, +∞) and there must be an asymptote at x = h The range is all real numbers Now find three points; the simplest values are when x - h = 1 and when x - h = b Graph of a Logarithmic Function Graph of y = logb(x) when b>1 • Graph of y = logb(x) when 0<b<1 Can you state any characteristics? –Asymptotes, x - intercepts, Domain, Range Practice Graph. State the domain and range. 11)y = log4 x 12)y = log 1 x - 3 2 13)y = log 5 (x - 2) 14)y = log 2 (x + 5) -1 Change of Base Formula log a c log b c = log a b This formula allows us to compute logs using the calculator, by converting to base 10 or e. Example: log(5) ln(5) log 3 5 = = log(3) ln(3)