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More notes on logs: log x represents a base 10 log , i.e. log x = log base is written in a log statement, then the base is 10. (Sort of like the radical with no index means square root...) ln x represents a base e log , i.e. ln x = log represent "natural log" or the log with the "natural" base... natural base. 10 x. Notice that if no e x. We use ln to e is that We looked at the graph of y = log b x when b > 1. We saw that the graph had a vertical asymptote at x = 0, contained the points (1, 0) and (b, 1). The graph showed increased from quadrant IV into quadrant I. Let's look at the graph of y = log b x when 0 < b < 1. Picture the graph of the inverse exponential decay function and reflect in the line y = x, switch all the "x" and "y" stuff to get the graph of this log function. Compare and contrast the graphs of these two basic logs, the one with b > 1 and the one with 0 < b < 1. 1 2 3 4 5 Properties of logs (because logs are exponents!!!!) log b (mn) = log exponenets... examples: b (n) because when you multiply numbers, you add their Write as the log of a single number: log 2 + log 3 log 2 + log 3 = log (2*3) = log 6 Write using logs of prime numbers only: log 24 log 24 = log (2*2*2*3) = log 2 + log 2 + log 2 + log 3 = 3log 2 + log 3 log b (m/n) = log exponents.... examples: b (m) + log b (m) log b (n) because when you divide numbers, you subtract their Write as the log of a single number: log 18 log 24 log 18 log 24 = log (18/24) = log (3/4) Simplify: log 50 + log 4 log 2 log 50 + log 4 log 2 = log (50*4/2) = log 100 = 2 If log 2 = a and log 3 = b, write each of the following in terms of a and/or b a) log 6: log 6 = log 2 + log 3 = a + b b) log (3/5) = log 3 log 5 = b log 5 = b log (10/2) = b (log 10 log 2) = b (1 a) = b 1 + a 6 log b (m n ) = n*log the exponents.... examples: b (m) because when you take the power of a power you multiply 3 log 8 = log 2 = 3 log 2 4 log 10000 = log 10 = 4 log 10 = 4 (reinforces the inverse relationship between logs and exponentials of the x property that log b b = x.) same base and the Combining the three properties: Simplify log 12 + 2log 4 3log 8 log 12 + 2log 4 3log 8 = log 12 + log 4 = log (12*4 2 /8 3 ) = log (3/8) If log 4 3 = m and log 4 5 = n, write log 4 3 = m and log log log 4 (12) 4 5 = n, write log 4 (1/12) = log 4 (4*3) = 1(log 4 (4*3) ) = 1(log 4 4 + log = 1(1 + m) = 1 m log 8 3 4 60 in terms of m and/or n. log 4 60 = log 4 (4*3*5) = log 4 4 + log 4 3 + log = 1 + m + n If log 2 1 4 5 4 (1/12) in terms of m and/or n. 4 3) OR log 4 (1/12) = log 4 (1) = 0 (log 4 4 + log = (1 + m) = 1 m 4 3) 7 8 9 10 Change of base rule. We do sometimes need to find decimal approximations for log expressions. We can estimate these values to the nearest integer but to get a better estimate we need to use a calculator. Our calculator has two log keys: log and ln. This implies we have to be able to write every log expression using base 10 or base e . We use the definition of logs to "get rid" of the base we don't want and then take the log using the base we do want to derive this rule. Example: Consider log 5 16 1 We know log 5 16 is somewhere between 1 and 2 (because 5 = 5 and 5 for the value of log 5 16 we need to switch this expression to one using base 10 or base 2 = 25). But if we want a better approximation e . Let's switch to base e. x log 5 16 = x means 5 = 16 by definition of log function ln (5 x ) = ln 16 take natural log of both sides of the equation x ln 5 = ln 16 using the "zoom" or power rule x = ln 16 ln 5 divide both sides by ln 5 Now we use the calculator to approximate a value for this expression. Notice where the 16 and the 5 in the original log function are in the new expression. Thus the change of base rule is log b m = log a m log ab where a is any new base you want 11 12 13