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Warm-up

Warm-up. 1. 2. 3. 4. Answers. 1. 2. 3. 4. About 148 decibels. Precalculus Lesson 3.4. Check: P. 241 #20, 22, 28, 32, 36, 44, 54, 58, 78, 84 . Quiz 3.3-3.4 tomorrow. Objective: Simplify expressions using properties of logs. Solve exponential and logarithmic equations. 2. 1.

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Warm-up

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  1. Warm-up 1. 2. 3. 4.

  2. Answers 1. 2. 3. 4. About 148 decibels

  3. PrecalculusLesson 3.4 Check: P. 241 #20, 22, 28, 32, 36, 44, 54, 58, 78, 84

  4. Quiz 3.3-3.4 tomorrow Objective: • Simplify expressions using properties of logs. • Solve exponential and logarithmic equations.

  5. 2. 1. 3. 4. 1. 0.222 2. 1.176 3. 1.398 4. 1.653

  6. Lesson Classwork 1: worksheet 7.5B Notes Powerpoint 3.4

  7. For b > 0 and b ≠ 1, if bx= by, then x = y. SOLVING EXPONENTIAL EQUATIONS If two powers with the same base are equal,then their exponents must be equal.

  8. Solving by Equating Exponents CHECK Solve 43x= 8x+ 1. 43x= 8x+1 SOLUTION Write original equation. (22)3x= (23)x +1 Rewrite each power with base 2. Check the solution by substituting it into the original equation. 22(3x)= 23(x +1) Power of a power property 43 • 1= 81+1 Solve for x. 26x= 23x + 3 64= 64 Solution checks. 6x= 3x + 3 Equate exponents. x= 1 Solve for x. The solution is 1.

  9. YOUR TURN #1 5 4x= 125 x+ 4 X=12

  10. Solving by Equating Exponents When it is not convenient to write each side of an exponential equation using the same base, you cansolve the equation by taking a logarithm of each side.

  11. Taking a Logarithm of Each Side 1 1 x = (3 + log 17) Multiply each side by . 2 2 Solve 102x – 3+ 4 = 21. SOLUTION 102x – 3+ 4 = 21 Write original equation. 102x – 3= 17 Subtract 4 from each side. log 102x – 3= log 17 Take common log of each side. 2x – 3 = log 17 log 10x = x 2x = 3 + log 17 Add 3 to each side. x≈2.115 Use a calculator.

  12. Taking a Logarithm of Each Side CHECK y x 1.0 2.0 Solve 102x – 3+ 4 = 21. Check the solution algebraically by substituting into theoriginal equation. Or, check it graphically by graphingboth sides of the equation and observing that the two graphs intersect at x≈2.115. y= 102x – 3+ 4 y= 21

  13. YOUR TURN #2 10 3 x – 1+ 5 = 25 X=0.767

  14. SOLVING LOGARITHMIC EQUATIONS To solve a logarithmic equation, use thisproperty for logarithms with the same base: For positive numbers b, x,andywhereb ≠ 1, Log b x = log b yif and only ifx = y.

  15. Solving a Logarithmic Equation CHECK ? log3(5 · 2 – 1) = log3(2 + 7) Solve log3(5x – 1) = log3(x + 7) . SOLUTION Check the solution by substituting it into the original equation. log3 (5x – 1) = log3 (x + 7) Write original equation. log3(5x – 1) = log3(x + 7) Write original equation. 5x – 1 = x + 7 Use property for logarithms with the same base. Substitute 2 for x. 5x = x + 8 Add 1 to each side. log39 = log39 Solution checks. x = 2 Solve for x. The solution is 2.

  16. Solving a Logarithmic Equation log5(3x + 1) 5 = 52 logbx b = x Solve log5(3x + 1) = 2 . SOLUTION log5(3x + 1) = 2 Write original equation. Exponentiate each side using base 5. 3x + 1 = 25 x= 8 Solve for x. The solution is 8.

  17. YOUR TURN #3 log 4 (7x + 1) = 3 . X=9

  18. Checking for Extraneous Solutions Because the domain of a logarithmic function generally does not include all real numbers, you should be sure to check for extraneous solutions of logarithmic equations. You can do this algebraically or graphically.

  19. Checking for Extraneous Solutions log (5x2 – 5x) 10 = 102 Check for extraneoussolutions. log 5x + log (x – 1) = 2 Solve log 5x + log (x – 1) = 2. SOLUTION Write original equation. log [5x (x – 1)] = 2 Product property of logarithms. Exponentiate each side using base 10. 5x2 – 5x =100 10logx = x x2–x–20 = 0 Write in standard form. (x – 5)(x + 4) = 0 Factor. x= 5 or x = –4 Zero product property

  20. Checking for Extraneous Solutions log 5x + log (x – 1) = 2 SOLUTION Check for extraneous solutions. x= 5 or x = –4 Zero product property y x The solutions appear to be 5 and –4. However, when you check these in the original equation or use a graphic check as shown below, you can see that x = 5 is the only solution. y= 2 y= log 5x + log (x – 1)

  21. YOUR TURN #4 log z + log (z + 3) = 1 Z = 2 -5 is extraneous

  22. Using a Logarithmic Model SEISMOLOGYOn May 22, 1960, a powerful earthquake took place in Chile. It had a moment magnitude of 9.5. How much energy did this earthquake release? The moment magnitude M of an earthquake that releases energy E (in ergs) can be modeled by this equation: M= 0.291 ln E + 1.17

  23. Using a Logarithmic Model SOLUTION M= 0.291 ln E + 1.17 Write model for moment magnitude. 9.5= 0.291 ln E + 1.17 Substitute 9.5 for M. 8.33 = 0.291 ln E Subtract 1.17 from each side. 28.625 ≈ ln E Divide each side by 0.291 e28.625≈ eln E Exponentiate each side using base e. 2.702 x 1012≈ E e ln x = e logex= x The earthquake released about 2.7 trillion ergs of energy.

  24. Assignments Classwork: p. 251 #15, 23, 37, 39, 59, 89, 99 Homework(3.4) p. 251 #14,16,18,56,70,86,88,102,108,110

  25. Closure Review properties of logs logaa=1 ln e =1 loga1=0 ln 1 = 0 logaax=x ln ex =x

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