Module 6 Lecture Notes Contents 6. An Introduction to Logarithms....................... 6. Evaluating Logarithmic Expressions.................... 4 6.3 Graphs of Logarithmic Functions...................... 6 6.4 Determining the Domain of a Logarithmic Function.......... 7 6.5 Solving Exponential and Logarithmic Equations............. 8 6.6 Applications to Logarithmic and Exponential Equations........ 0 6.7 Properties of Logarithms........................... 6.8 The Change-of-Base Formula........................ 8 6.9 Summary of the Properties of Logarithms................ 9 6. An Introduction to Logarithms We looked at equations such as 8 x = x+ in the last section, and solved them by rewriting the lefthand side with a base of. But what if we have equations such as x = 0? We know intuitively that 3 < x < 4, but can we give an exact answer? We will need functions that are inverse functions to exponential functions to solve such equations. The logarithmic function to the base a, where a > 0 and a, is denoted by y = log a x) and is defined by y = log a x) if and only if x = a y
The common logarithmic function is the logarithmic function with base 0 given by fx) = logx). We write logx) to represent log 0 x) The natural logarithmic function is the logarithmic function with base e given by fx) = lnx). We write lnx) to represent log e x) Example : Solve the equation x = 0 by converting from exponential form to logarithmic form. x = 0 x = log 0) Notice in the example above that the base was the same in each the exponential equation and the logarithmic equation. The base logarithmic function is the inverse of the base exponential function. In the equations above, the base stays the same. When changing from exponential form to logarithmic form, the x and the 0 changed places as one would expect to happen with inverses. Example : Change the exponential statement to an equivalent statement using logarithms. a) ) x 4 = 6 b) 0 m = 7 ) x = 6 4 log /4 6) = x 0 m = 7 log 0 7) = m log7) = m c) e t = 5 Instructor: A.E.Cary Page of 9
e t = 5 log e 5) = t If we continued solving for t : ln5) = t ln5) = t Example 3: Change the logarithmic statement to an equivalent statement using exponents. a) log 3 5) = x b) logx) = c) lnx) = 5 log 3 5) = x 3 x = 5 logx) = 0 = x lnx) = 5 e 5 = x d) log y + ) = 3 log y + ) = 3 3 = y + If we continue solving for y : 8 = y + 8 = y 7 8 = y Instructor: A.E.Cary Page 3 of 9
6. Evaluating Logarithmic Expressions Example 4: Find the exact value of each logarithmic expression without using a calculator. a) log 8) log 8) = log 3 ) = 3 b) log 3 3 ) log 3 3 ) = c) lne 3 ) lne 3 ) = 3 d) log 5 5) log 5 5) = log 5 5 ) = e) log ) 4 64 ) = log 4 4 3 ) log 4 64 = log 4 4 3 ) = 3 f) ln ) e ln ) = ln e ) e = Instructor: A.E.Cary Page 4 of 9
g) log00) log00) = log 0 0 ) = h) log 4 ) log 4 ) = log 4 4 0 ) = 0 i) log 6 6 ) ) log 6 6 = log ) 6 6 / = Instructor: A.E.Cary Page 5 of 9
6.3 Graphs of Logarithmic Functions Example 5: If fx) = x, then the inverse function of f is given by f x) = log x). We can sketch the graph of y = fx) by creating a table of values, as shown in Table 5 and Figure 6.. Table 6. x x -4 6-3 8-4 - 0 4 3 8 y = 0 y = fx) Figure 6. y 8 6 4 8 6 4 4 6 8 4 6 8 x To graph y = f x) and create a table of values, we can use the properties of inverses functions. Since the point 3, 8) is on the graph of y = fx), the point, 3) will be on the graph of 8 y = f x. Likewise, the points, ),, ),, 0),, ), and 4, ) and 8, 3) will all be on the 4 graph of y = f x). This is shown in Figure 6. and Table 5. Table 6. x log x) 6-4 8-3 4 - - 0 4 8 3 y = x y = 0 x = 0 y = fx) y = f x) Figure 6. y 8 6 4 8 6 4 4 6 8 4 6 8 x Instructor: A.E.Cary Page 6 of 9
Based on the graphs in the previous example, we can see the domain, range, and any asymptotes for each function, which are summarized below: Table 6.3 Function Domain Range f, ) 0, ) f 0, ), ) 6.4 Determining the Domain of a Logarithmic Function Example 6: State the domain of the following functions: a) fx) = log 5 x) Since a logarithmic function can only be evaluated at numbers greater than zero, the domain of f is {x x > 0}. b) gx) = log 4 x + 5) Since a logarithmic function can only be evaluated at numbers greater than zero, the inequality x + 5 > 0 must hold. Thus x > 5 and so the domain of g is {x x > 5}. c) hx) = log 9 7 x) Since a logarithmic function can only be evaluated at numbers greater than zero, the inequality 7 x > 0 must hold. Thus x < 7 and so the domain of h is {x x < 7}. d) jx) = log ) x 3 This domain of j is a bit more challenging to determine. We need > 0, which cannot x 3 be solved algebraically. If we solve this numerically or graphically, we see that > 0 for x 3 x > 3. Thus the domain of j is {x x > 3}. Instructor: A.E.Cary Page 7 of 9
6.5 Solving Exponential and Logarithmic Equations We can use the properties of logarithms to solve equations containing exponential expressions. Logarithms are necessary to solve equations where the variable is in the exponent and each side of the equation does not have a common base). If the variable we are solving for is inside a logarithm, we will need to convert this to exponential form in order to solve for the varaible. Example 7: Solve the following equations for x and state the solution set. a) e x+ = 3 Solution set: { ln3) } e x+ = 3 x + = ln3) x = ln3) x = ln3) x = ln3) b) 8 0 7x = 4 Solution set: { } log ) 7 8 0 7x = 4 8 0 7x 8 = 4 8 0 7x = ) log = 7x log ) = x 7 Instructor: A.E.Cary Page 8 of 9
c) log 3 5x 4) = Solution set: { } 3 5 log 3 5x 4) = 3 = 5x 4 9 = 5x 4 3 = 5x 3 5 = x d) log x + ) = 3 log x + ) = 3 3 = x + 8 = x + 7 = x ± 7 = x Solution set: { 7, 7 } Instructor: A.E.Cary Page 9 of 9
6.6 Applications to Logarithmic and Exponential Equations Example 8: Let ph) be the atmospheric pressure on an object measured in millimeters of mercury) that is h kilometers above sea level. The function p can be modeled by ph) = 760e 0.45h Find the height of a mountain where the atmospheric pressure is 60 millimeters of mercury. To determine the height of a mountain where the atmospheric pressure is 60 millimeters of mercury, we need to solve ph) = 60: 60 = 760e 0.45h 60 760 = e 0.45h 3 38 = e 0.45h ) 3 ln = 0.45h 38 ln ) 3 38 0.45 = 0.45h 0.45 ln ) 3 38 0.45 = h h.404 The height of the mountain is approximately.404 km or,404 meters). Instructor: A.E.Cary Page 0 of 9
Example 9: The ph of a chemical solution is given by the formula ph = log 0 [ H + ] where [ H +] is the concentration of hydrogen ions in moles per liter. Values of ph range from 0 acidic) to 4 alkaline). a) What is the ph of a solution for which the concentration of hydrogen ions [ H +] ) is 0.0 moles per liter? Since we are given that [ H +] = 0.0, we will replace [ H +] in the equation above to determine the ph: ph = log 0 0.0) ) = log 0 00 = log ) 0 0 = ) = The ph of a solution for which the concentration of hydrogen ions [ H +] ) is 0.0 moles per liter is. b) What is the concentration of hydrogen ions [ H +] ) in a banana with a ph of 4.5? Since we are given that ph = 4.5, we will replace ph in the equation above and solve for [ H + ] : [ 4.5 = log ] 0 H + [ 4.5 = log ] 0 H + [ ] H + = 0 4.5 [ H + ] 0.00036 The concentration of a banana is about 0.0036 moles per liter. Instructor: A.E.Cary Page of 9
6.7 Properties of Logarithms For each rule of exponents, there exists a logarithmic counterpart. For example, because a 0 =, it holds that log a ) = 0. Similarly, as a = a, log a a) =. If we think about these with concrete examples, this result should seem logical: log 5 ) = log 5 5 0 ) = 0 log 5 5) = log 5 5 ) = For any positive real number a, a, it holds that log a ) = 0 log a a) = Since fx) = a x and gx) = log a x), we know that fgx)) = x and gfx)) = x. But what do fgx)) and gfx)) look like? fgx)) = flog a x)) = a log a x) gfx)) = ga x ) = log a a x ) For any positive real numbers x and a, a, it holds that log a a x ) = x a log a x) = x Example 0: Use the properties of logarithms to find the exact value of the following expressions without using a calculator. a) log 4 4 5 ) log 4 4 5 ) = 5 b) log 0 7) log 0 7) = 7 Instructor: A.E.Cary Page of 9
c) e ln) e ln) = d) log ) log ) = Exponent Rules: a m+n = a m a n a m n = am a n a m) n = a mn For each of the exponent rules above, we have logarithmic counterparts. As you work through applying these properties, it s good to write them down each time you use them. You will need to know all of the main properites of logarithms, and the best way to commit them to memory is to get a lot of practice with them. At the end of these lecture notes, you will find a complete list of the properties. Properties of Logarithms: For any positive real numbers M, N, and a, a, it holds that log a MN) = log a M) + log a N) ) M log a = log N a M) log a N) log a M r ) = r log a M) Instructor: A.E.Cary Page 3 of 9
Example : Use the properties of logarithms to find the exact value of the following expressions. Do not use a calculator. a) log 3 6) log 3 ) ) 6 log 3 6) log 3 ) = log 3 = log 3 3) = b) log 8 6) + log 8 4) log 8 6) + log 8 4) = log 8 6 4) = log 8 64) = log 8 8 ) = 64 c) 4 log 4 7)+log 4 8) 4 log 4 7)+log 4 8) = 4 log 4 7 8) = 4 log 4 56) = 56 Alternately, we could do the following: 4 log 4 7)+log 4 8) = 4 log 4 7) 4 log 4 8) = 7 8 = 56 Instructor: A.E.Cary Page 4 of 9
Example : Write each expression as a sum and/or difference of logarithms. Express powers as factors. ) a) log, x > 3 x 3 ) log = log) logx 3) By Prop. 6 x 3 = 0 logx 3) By Prop. = logx 3) b) log 3 x 4 5 + x ) log 3 x 4 5 ) + x = log ) ) 3 x 4 5 + log 3 + x By Prop. 5 = log 3 x 4 ) + log 3 + x ) /5) = 4 log 3 x) + 5 log 3 + x ) By Prop. 7 c) ln ) x + 4 x 4 ln ) x + 4 ) = ln x + 4 lnx 4) By Prop. 6 x 4 x = ln + 4 ) ) / ln x )x + ) ) = lnx + 4) [ lnx ) + lnx + ) ] By Prop. 7 and Prop. 6 = lnx + 4) lnx ) lnx + ) Instructor: A.E.Cary Page 5 of 9
Example 3: Write each expression as a single logarithm. ) ) a) log 4 + log x 3 4 x 4 ) ) log 4 + log x 3 4 x 4 = log 4 x ) 3 x 4 = log 4 x 7 ) or 7 log 4 x) By Prop. 5 ) ) b) log 4 log x 3 4 x 4 ) ) log 4 log x 3 4 = log x 4 4 x ) 3 x 4 ) = log 4 x x4 3 = log 4 x) By Prop. 6 c) log x + 5x + 6 ) log x + 3) log x + 5x + 6 ) log x + 3) = log x + 5x + 6 ) log x + 3) ) By Prop. 7 = log x + 3)x + ) x + 3) ) By Prop. 6 ) x + 3)x + ) = log x + 3) ) x + 3)x + ) = log x + 3) = log ) x + x + 3 Instructor: A.E.Cary Page 6 of 9
d) ln ) ) x + 9x + 4 x 49 + ln x 6x 7 x + x ln ) ) ) x + 9x + 4 x 49 x + 9x + 4 + ln = ln x 6x 7 x + x x 6x 7 x 49 x + x ) x + 7)x + ) x 7)x + 7) = ln x 7)x + ) xx + ) ) x + 7) x + ) = ln xx + ) Instructor: A.E.Cary Page 7 of 9
6.8 The Change-of-Base Formula If a, b, and M are positive real numbers, a and b, then log a M) = log bm) log b a) In practice, we primarily use one of the following forms of this formula: log a M) = logm) loga) or log a M) = lnm) lna) Example 4: Use the Change-of-Base formula to write the following logarithmic expressions in terms of the natural logarithmic function or common logarithmic function. Then approximate each in your calculator. a) log 5) log 5) = log5) and log log) 5) = ln5) ln) log 5).3 b) log 5 ) log 5 ) = log ) log5) and log 5 ) = ln ) ln5) log 5 ) 0.43 Instructor: A.E.Cary Page 8 of 9
6.9 Summary of the Properties of Logarithms If a, b, M and N are positive real numbers, a and b, then log a ) = 0 log a a) = log a a x ) = x a log a x) = x log a MN) = log a M) + log a N) ) M log a = log N a M) log a N) log a M r ) = r log a M) log a M) = log bm) log b a) Instructor: A.E.Cary Page 9 of 9