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Lesson 11 Logarithm and Common Logarithm

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Lesson 11 Logarithm and Common Logarithm Course I Lesson １１ Logarithm and Common Logarithm １１A • Logarithm of a number • Merits of logarithm • Logarithmic Identities 1 Logarithm of a Number M = a p Definition The logarithm of a number M is the exponent p by which another fixed value (the base) a has to be raised to produce that number, and it written as p = log a M . y In other words 3 2 = log2 4 ○ The value p is the answer to the 2 y = log2 x question "To what power must a 1 O 2 4 6 8 10 x be raised in order to yield M ?". −1 −2 −3 −4 2 Merits of Logarithm (1) Merit (1) • Logarithm can represent a large number by a small number. Ex. x =100000 → log x = 5 10 • Logarithmic scale is helpful to represent wide range of values with different details. (See next slide for illustration) That makes sense! 3 Logarithmic Scale http://www.isas.jaxa.jp/home/solar/home/ solar/yohkoh/Graph/Yohkoh full.gif ６ 1000000 1×10 ℃ Corona 100000 10000 ３ 6×10 ℃ Surface of the sun 1000 Logarithmic scale 100 100 ℃ Boiling water 40 ℃ Bath 10 4 1 Logarithmic Identities Product loga MN = loga M + loga N M Quotient log = log M − log N a N a a k Power loga M = k loga M [ Example 11.1] Prove the product rule of logarithm. Ans. p q Let M = a and N = a . Then p = loga M, q = loga N By multiplication M ⋅ N = a p⋅ aq = a p+q Therefore loga M ⋅ N = p + q = loga M + loga N 5 Special Cases Special Values 1 a = a loga a =1 0 a = 1 loga 1 = 0 −1 1 ⎛ 1 ⎞ a = loga ⎜ ⎟ = −1 a ⎝ a ⎠ Special Identities M M =1 1 log = log M − log N loga = −loga N a N a a N 1 k = n k n 1 loga M = k loga M log M = log M a n a 6 Changing the Base of Logarithm [Example 11.2] Prove the following formula : log b log b = c a( 1), b( 1), c( 1) are positive real numbers a ≠ ≠ ≠ logc a Ans. p Let p = log a b , then b = a holds. If we make logarithms with base c of both sides, we have p logc b = logc a ∴ logc b = plogc a Since a ≠ 1 , we see logc a ≠ 0 log b Therefore, p = c logc a 7 Logarithmic Equation [Example 11.3] Solve the following equations. (1) log 2 x = 3 (2) log2(x + 3) + log2(2x −1) = 2 Ans. 3 (1) From the definition x = 2 = 8 (2) Because x + 3 > 0 and 2x −1 > 0 1 Therefore x > 2 The given expression becomes log2(x +3)(2x −1) = log2 4 ∴ (x + 3)(2x −1) = 4 ∴ 2x2 + 5x − 7 = 0 ∴ (x −1)(2x + 7) = 0 7 ∴ x =1, x = − 2 1 Since x > , we have x = 1 2 8 Logarithmic Inequality [Example 11.4] Solve the following inequality. log x < 3 2 Ans. 3 log2 x < 3 = log2 2 = log2 8 From the graph of y = log 2 x , we have 0 < x < 8 y 3 y = 3 O 1 8 x y = log2 x 9 Exercise [Ex.11.1] Find the values of (1) log 5 125 (2) log 10 10 10 [Ex.11.2] Simplify the following expressions: (1) log 10 2 + log 10 5 (2) log 3 7 − log3 63 10 Answer to the Exercise [Ex.11.1] Find the values of (1) log 5 125 (2) log 10 10 10 (1) We put log 5 125 = p . , Then, we have 5p =125 = 53 ∴ p = 3 (2) 3 3 3 log 10 10 = log 10 2 = log 10 = 10 10 2 10 2 . [Ex.11.2] Simplify the following expressions: (1) log 10 2 + log 10 5 (2) log 3 7 − log3 63 (1) log10 2 + log10 5 = log10 (2 + 5)= log10 10 =1 7 1 −2 (2) log3 7 − log3 63 = log3 = log3 = log3 3 = −2 63 9 11 Exercise [Ex.11.3] (1) Simplify the following expression (log2 9 + log4 3)(log3 2 + log9 4) (Hint: Make a common base.) (2) Solve the following logarithmic equation log2 x +log2 (x −7) = 3 Pause the video and solve the problem by yourself. 12 Answer to the Exercise [Ex.11.3] (1) Simplify the following expression (Hint: Make a common base.) (2) Solve the following equation. log2 x +log2 (x −7) = 3 ⎛ log 3 ⎞⎛ log 2 log 4 ⎞ log 9 log 3 log 2 log 4 log 9 2 2 2 (1) ( 2 + 4 )( 3 + 9 ) = ⎜ 2 + ⎟⎜ + ⎟ ⎝ log2 4 ⎠⎝ log2 3 log2 9 ⎠ log 3 ⎛ 1 2 ⎞ ⎛2log 3 2 ⎞ = ⎜ 2 + ⎟⎜ + ⎟ ⎝ 2 ⎠⎝ log2 3 2log2 3⎠ 5 2 = log2 3⋅ = 5 2 log2 3 (2) log2 x +log2 (x −7) = 3 Since, x > 0 and x − 7 > 0 , therefore x > 7 3 log2 x(x − 7) = log2 2 ∴ x(x − 7) = 8 ∴ (x +1)(x − 8) = 0 From ①, the answer is x = 8 13 Course I Lesson １１ Logarithm and Common Logarithm １１B • Special Logarithm • Common Logarithm 14 Various Bases Various exponential function Various logarithmic function x 2 x 5 y = log x ⎛ ⎞ y ⎛ ⎞ 3 y = ⎜ ⎟ = ⎜ ⎟ 2 ⎝ 5 ⎠ y ⎝ 2 ⎠ y y = log2 x 3 x 8 ⎛ 1 ⎞ x y = ⎜ ⎟ y = 2 2 ⎝ 2 ⎠ y = log5 x 6 2 1 4 O x x 2 4 6 8 x ⎛ 2 ⎞ ⎛ 3 ⎞ y = ⎜ ⎟ y = ⎜ ⎟ −1 ⎝ 3 ⎠ 2 ⎝ 2 ⎠ −2 y = log2 x 5 −3 −2 −1 O 1 2 3 x −3 y = log1 x y = log2 x 2 3 Then, which one shall we use ? → We are familiar to decimal number. 2356 = 2×103 + 3×10 2 + 5×10 + 6 15 Special Logarithms Common Logarithm a = 10 y = log10 x or y = log x • This has applications in science and engineering. Natural Logarithm a = e = 2.718... y = loge x or y = ln x • This is also written as y = ln x • This is used in pure mathematics. • This will be explained later in the lesson on differentiation. Binary Logarithm a = 2 y = log2 x • This is used in computer science. 16 Common Logarithm • Values of common logarithm are given in the logarithmic table. • The table gives the value for 1< M <9.99. • For M = a × 10 n ( 1 ≤ a < 10 ) , the logarithmic value is obtained by n log10 M = log10 (a×10 ) = n + log10 a Table of Common Logarithm 17 Examples [Example 11.5] Find the common logarithm of the following numbers. 0.01, 0.1, 1, 10, 100 Ans. −2 log10 0.01 = log10 10 = −2 log10 10 =1 log10 0.1= −1 log10 100 = 2 log10 1= 0 [Example 11.6] Find the common logarithm of 11910 from the logarithmic table. Ans. 4 log10 11910 = log10 (1.191×10 ) ≈ 4 + log10 1.19 = 4 + 0.0755 = 4.0755 18 Examples [Example 11.7] Find the number of digits of 625 . Use the following values. log102 = 0.3010, log10 3 = 0.4771 Ans. 25 Put N = 6 and take the common logarithm of both sides. 25 log10 N = log10 6 = 25×(log102 + log10 3) = 25×(0.3010 + 0.4771) =19.4525 Then N =1019.4525 Therefore 1019 < N <10 20 This means that 6 25 is a 20 digits number . 19 Merit of Logarithm (2) By this product rule log a MN = log a M + log a N , we can replace the laborious product by an easy addition. [ Example 11.8] x = 123111 , y = 9663189 are given. Find the approximate value of xy using a logarithmic table in the previous slide. Ans. 5 5 x =1.23111×10 ∴ log10 x = log10 1.23111+ log10 10 ≈ log10 1.23 + 5 6 6 y = 9.663189×10 ∴ log10 y =log10 9.663189 + log10 10 = log10 9.66 + 6 From table log101.23 = 0.0899, log10 9.66 =0.9850 Therefore log10 xy = log10 x + log10 y ≈ (log10 1.23 + 5) + (log10 9.66 + 6) 12 ≈ 5.0899 + 6.9850 ≈12 + 0.0749 ≈12 + log10 1.19 = log10 (1.19 ×10 ) xy 1.19 1012 ∴ ≈ × 20 Slide Rule y Log(y) Log(x) x Log(xy) xy 2×3=6 Fig.1 Principle Fig.2 Overview 21 Example [Example 11.7] When we keep 1 million yen in the bank with the rate of interest 5%, how much can we get after 5 years? Use the table in the former slide. Ans. y =1000000 ×1.055 5 log10 y = 6 + log10 1.05 = 6 + 5× log10 1.05 = 6 + 5× 0.0212 = 6 + 0.106 From the table 0.106 ≈ log10 1.28 6 4 log10 y ≈ 6 + log10 1.28 = log10 (1.28 ×10 ) = log10 (128 ×10 ) We can get about 1.28 million yen The precise value is 1000000 ×1.055 =1276281.5 22 Exercise [Ex.11.4] Let log 10 2 = 0 . 3010 and log 10 3 = 0 . 4771 . Find the values of (1) log10 125, (2) log2 5 Pause the video and solve the problem by yourself. 23 Answer to the Exercise [Ex.11.4] Let log 10 2 = 0 . 3010 and log 10 3 = 0 . 4771 . Find the value of (1) log10 125, (2) log2 5 (1) 3 log10 125 = log10 5 = 3log10 5 = 3log10 (10 / 2) = 3(1− log10 2) = 3×(1− 0.3010)= 2.0970 (2) log10 5 1− log10 2 1− 0.3010 log2 5 = = = ≈ 2.322 log10 2 log10 2 0.3010 24 .
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