Section 2.6 Logarithmic Functions
Section 2.6 Logarithmic Functions One-to-One Functions HORIZONTAL LINE TEST: A function is one-to-one if and only if no horizontal line intersects its graph more than once. DEFINITION: Let f be a one-to-one function with domain A and range B. Then its inverse function f −1 has domain B and range A and is defined by f −1 (y) = x ⇐⇒ f (x) = y (∗) for any y in B. So, we can reformulate (∗) as f −1 (f (x)) = x for every x in the domain of f f (f −1 (x)) = x for every x in the domain of f −1 IMPORTANT: Do not confuse f −1 with 1 . f EXAMPLES: 1. Let f (x) = x3 , then f −1 (x) = √ 3 x, since √ √ 3 f −1 (f (x)) = x3 = x and f (f −1 (x)) = ( 3 x)3 = x 1 2. Let f (x) = 5x, then f −1 (x) = x, since 5 f −1 1 (f (x)) = (5x) = x and f (f −1 (x)) = 5 5 1 1 x 5 =x THEOREM: If f has an inverse function f −1 , then the graphs of y = f (x) and y = f −1 (x) are reflections of one another about the line y = x; that is, each is the mirror image of the other with respect to that line. DEFINITION: Let a be a positive number with a 6= 1. The logarithmic function with base a, denoted by loga , is defined by loga x = y ay = x ⇐⇒ So, loga x is the exponent to which the base a must be raised to give x. 4 4 4 y 2 y 2 y 2 0 -4 -2 0 0 2 4 -4 -2 0 0 2 x 4 0 2 x 4 6 8 10 x -2 -2 -2 -4 -4 -4 BASIC PROPERTIES: f (x) = loga x is a continuous function with domain (0, ∞) and range (−∞, ∞). Moreover, loga (ax ) = x for every x ∈ R, aloga x = x for every x > 0 REMARK: It immediately follows from property 1 that loga a = 1, loga 1 = 0 EXAMPLES: 1. log2 2 = 1, log2 4 = 2, log2 8 = 3, log2 16 = 4 2. log3 3 = 1, log3 9 = 2, log3 27 = 3, log3 81 = 4 1 1 1 −1 −2 3. log3 = log3 3 = −1, log3 = log3 3 = −2, log3 = log3 3−3 = −3 3 9 27 2 √ 1 1 3 5 = log5 51/2 = , log7 7 = log7 71/3 = 2 3 1 1 1 = log11 = log11 11−1/5 = − 5. log11 √ 5 1/5 11 5 11 4. log5 √ Common and Natural Logarithms DEFINITION: The logarithm with base e is called the natural logarithm and has a special notation: loge x = ln x BASIC PROPERTIES: 1. ln(ex ) = x for every x ∈ R. 2. eln x = x for every x > 0. REMARK: It immediately follows from property 1 that ln e = 1 DEFINITION: The logarithm with base 10 is called the common logarithm and has a special notation: log10 x = log x BASIC PROPERTIES: 1. log(10x ) = x for every x ∈ R. 2. 10log x = x for every x > 0. REMARK: It immediately follows from property 1 that log 10 = 1 3 Laws of Logarithms LAWS OF LOGARITHMS: If x and y are positive numbers, then 1. loga (xy) = loga x + loga y. x = loga x − loga y. 2. loga y 3. loga (xr ) = r loga x where r is any real number. EXAMPLES: 1. Use the laws of logarithms to evaluate log2 4. Solution: We have log2 4 = log2 22 = 2 log2 2 = 2 · 1 = 2 2. Use the laws of logarithms to evaluate log10 1000000. Solution: We have log10 1000000 = log10 106 = 6 log10 10 = 6 · 1 = 6 3. Use the laws of logarithms to evaluate log7 Solution: We have log7 √ √ 7 = log7 71/2 = 7. 1 1 1 log7 7 = · 1 = 2 2 2 1 4. Use the laws of logarithms to evaluate log5 √ . 3 5 Solution: We have 1 1 1 1 1 = log5 1/3 = log5 5−1/3 = − log5 5 = − · 1 = − log5 √ 3 5 3 3 3 5 or 1 1 1 1 1 1 log5 √ = log5 1/3 = log5 1 − log5 51/3 = 0 − log5 5 = − log5 5 = − · 1 = − 3 5 3 3 3 3 5 5. Use the laws of logarithms to evaluate log3 270 − log3 10. Solution: We have log3 270 − log3 10 = log3 270 10 = log3 27 = log3 33 = 3 log3 3 = 3 · 1 = 3 6. Use the laws of logarithms to evaluate log2 12 + log2 3 − log2 9. Solution: We have log2 12 + log2 3 − log2 9 = log2 (12 · 3) − log2 9 = log2 12 · 3 9 = log2 4 = log2 22 = 2 log2 2 = 2 · 1 = 2 4 EXAMPLES: 1. ln (x(x + 1)) = ln x + ln(x + 1) x 2. ln = ln x − ln(x + 2) x+2 x(x + 1) = ln x(x + 1) − ln(x + 2) = ln x + ln(x + 1) − ln(x + 2) 3. ln x+2 x(x + 1) = ln x(x + 1) − ln (x + 2)(x + 3) 4. ln (x + 2)(x + 3) = ln x + ln(x + 1) − ln(x + 2) + ln(x + 3) = ln x + ln(x + 1) − ln(x + 2) − ln(x + 3) √ x x+1 x(x + 1)1/2 5. ln √ = ln 3 (x + 2)1/3 (x + 3)5 x + 2(x + 3)5 = ln x(x + 1)1/2 − ln (x + 2)1/3 (x + 3)5 = ln x + ln(x + 1)1/2 − ln(x + 2)1/3 + ln(x + 3)5 = ln x + ln(x + 1)1/2 − ln(x + 2)1/3 − ln(x + 3)5 = ln x + 1 1 ln(x + 1) − ln(x + 2) − 5 ln(x + 3) 2 3 Change of Base IMPORTANT FORMULA: For any positive a and b (a, b 6= 1) we have logb x = loga x loga b In particular, if a = e or 10, then logb x = EXAMPLE: log4 8 = log22 23 = log x ln x = ln b log b 3 log2 2 3·1 3 log2 23 = = = 2 log2 2 2 log2 2 2·1 2 5 Exponential Equations An exponential equation is one in which the variable occurs in the exponent. EXAMPLE: Solve the equation 2x = 7. Solution 1: We have 2x = 7 log2 2x = log2 7 x log2 2 = log2 7 x = log2 7 ≈ 2.807 Solution 2: We have 2x = 7 ln 2x = ln 7 x ln 2 = ln 7 x= ln 7 ≈ 2.807 ln 2 EXAMPLE: Solve the equation 4x+1 = 3. Solution 1: We have 4x+1 = 3 log4 4x+1 = log4 3 (x + 1) log4 4 = log4 3 x + 1 = log4 3 x = log4 3 − 1 Solution 2: We have 4x+1 = 3 ln 4x+1 = ln 3 (x + 1) ln 4 = ln 3 x+1= x= EXAMPLE: Solve the equation 3x−3 = 5. 6 ln 3 ln 4 ln 3 −1 ln 4 EXAMPLE: Solve the equation 3x−3 = 5. Solution 1: We have 3x−3 = 5 x − 3 = log3 5 x = log3 5 + 3 Solution 2: We have 3x−3 = 5 ln 3x−3 = ln 5 (x − 3) ln 3 = ln 5 x−3= x= ln 5 ln 3 ln 5 +3 ln 3 EXAMPLE: Solve the equation 8e2x = 20. Solution: We have 8e2x = 20 e2x = 20 5 = 8 2 2x = ln x= 5 2 ln 25 1 5 = ln 2 2 2 EXAMPLE: Solve the equation e3−2x = 4. Solution: We have e3−2x = 4 3 − 2x = ln 4 −2x = ln 4 − 3 x= 3 − ln 4 ln 4 − 3 = ≈ 0.807 −2 2 7 Logarithmic Equations A logarithmic equation is one in which a logarithm of the variable occurs. EXAMPLE: Solve the equation ln x = 8. Solution: We have ln x = 8 eln x = e8 x = e8 EXAMPLE: Solve the equation log2 (x + 2) = 5. Solution: We have log2 (x + 2) = 5 2log2 (x+2) = 25 x + 2 = 25 x = 25 − 2 = 30 EXAMPLE: Solve the equation log7 (25 − x) = 3. Solution: We have log7 (25 − x) = 3 7log7 (25−x) = 73 25 − x = 73 x = 25 − 73 = −318 EXAMPLE: Solve the equation 4 + 3 log(2x) = 16. 8 EXAMPLE: Solve the equation 4 + 3 log(2x) = 16. Solution: We have 4 + 3 log(2x) = 16 3 log(2x) = 12 log(2x) = 4 2x = 104 x= 104 = 5000 2 EXAMPLE: Solve the equation log(x + 2) + log(x − 1) = 1. Solution: We have log(x + 2) + log(x − 1) = 1 log[(x + 2)(x − 1)] = 1 (x + 2)(x − 1) = 10 x2 + x − 2 = 10 x2 + x − 12 = 0 (x + 4)(x − 3) = 0 x+4=0 or x = −4 x−3=0 x=3 We check these potential solutions in the original equation and find that x = −4 is not a solution (because logarithms of negative numbers are undefined), but x = 3 is a solution. EXAMPLE: Solve the following equations (a) log(x + 8) + log(x − 1) = 1 (b) log(x2 − 1) − log(x + 1) = 3 9 EXAMPLE: Solve the following equations (a) log(x + 8) + log(x − 1) = 1 (b) log(x2 − 1) − log(x + 1) = 3 Solution: (a) We have log(x + 8) + log(x − 1) = 1 log[(x + 8)(x − 1)] = 1 (x + 8)(x − 1) = 10 x2 + 7x − 8 = 10 x2 + 7x − 18 = 0 (x + 9)(x − 2) = 0 x+9=0 or x−2=0 x = −9 x=2 We check these potential solutions in the original equation and find that x = −9 is not a solution (because logarithms of negative numbers are undefined), but x = 2 is a solution. (b) We have log(x2 − 1) − log(x + 1) = 3 log x2 − 1 =3 x+1 x2 − 1 = 103 x+1 (x − 1)(x + 1) = 1000 x+1 x − 1 = 1000 x = 1001 EXAMPLE: Find the solution of the equation, correct to two decimal places. (a) 10x+3 = 62x (b) 5 ln(3 − x) = 4 (c) log2 (x + 2) + log2 (x − 1) = 2 10 EXAMPLE: Find the solution of the equation, correct to two decimal places. (a) 10x+3 = 62x (b) 5 ln(3 − x) = 4 (c) log2 (x + 2) + log2 (x − 1) = 2 Solution: (a) We have 10x+3 = 62x ln 10x+3 = ln 62x (x + 3) ln 10 = 2x ln 6 x ln 10 + 3 ln 10 = 2x ln 6 x ln 10 − 2x ln 6 = −3 ln 10 x(ln 10 − 2 ln 6) = −3 ln 10 x= −3 ln 10 ≈ 5.39 ln 10 − 2 ln 6 (b) We have 5 ln(3 − x) = 4 ln(3 − x) = 4 5 3 − x = e4/5 x = 3 − e4/5 ≈ 0.77 (c) We have log2 (x + 2) + log2 (x − 1) = 2 log2 (x + 2)(x − 1) = 2 (x + 2)(x − 1) = 4 x2 + x − 2 = 4 x2 + x − 6 = 0 (x − 2)(x + 3) = 0 x−2=0 or x=2 x+3=0 x = −3 Since x = −3 is not from the domain of log2 (x + 2) + log2 (x − 1), the only answer is x = 2. 11
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