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7 6 Practice Continued for Use With Pages 512 522 Slader

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Lesson 7.6 , For use with pages 515-522 PowerPoint Presentation

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Lesson 7.6 , For use with pages 515-522

WARMUP. –5, 12. ANSWER. ANSWER. ANSWER. 3 4 = 2 x – 7. 3. 2. 1. log 8 30 = x. ANSWER. ANSWER. 125. Lesson 7.6 , For use with pages 515-522. 1. Write log 3 (2 x – 7) = 4 in exponential form. Write 8 x = 30 in logarithmic form. Hint: reverse loop swoop doop.

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Lesson 7.6 , For use with pages 515-522

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  1. WARMUP –5, 12 ANSWER ANSWER ANSWER 34 = 2x – 7 3 2 1 log8 30 = x ANSWER ANSWER 125 Lesson 7.6, For use with pages 515-522 1.Write log3(2x – 7) = 4 in exponential form. • Write 8x= 30 in logarithmic form. • Hint: reverse loop swoop doop Solve for x. 3. 100x= 1000 4.log5x = –3 • x2 – 7x – 60 = 0 • Hint: Use the quadratic formula.

  2. 7.6 Notes - Solve Exponential and Log Equations https://www.youtube.com/watch?v=XjzBpJ1XYwc

  3. Rewrite each base as a power of 2. Distribute. Powers equal to each other. Objective - To solve exponential and logarithmic equations.

  4. Take the log3 of each side.

  5. Take the log5 of each side.

  6. If logb x = logb y if and only if x = y.

  7. Loop Swoop Doop

  8. You do this one If x = y, then bx = by. You probably did loop-swoop-doop, but here is another approach.

  9. 10 Check! Remember, you can't take the log of a negative!

  10. 1 x –3 x 2 Solve 4= x –3 2 x – x +3 x 2 = 2 1 1 4= Rewrite 4 and as powers with base 2. 2 2 x x –3 2 –1 (2 ) = (2 ) ANSWER The solution is 1. EXAMPLE 1 Solve by equating exponents SOLUTION Write original equation. Power of a power property Property of equality for exponential equations 2x = –x + 3 x = 1 Solve for x.

  11. –2 ? 4 = 1 1 2 2 1–3 ? 1 4 = EXAMPLE 1 Solve by equating exponents Check: Check the solution by substituting it into the original equation. Substitute 1 for x. Simplify. Solution checks. 4 = 4

  12. 1 – 8 3 5 5x –6 3 – x 3. 81 = x –1 2x 1. 9 = 27 7x +1 3x –2 2. 100 = 1000 for Example 1 GUIDED PRACTICE Solve the equation. SOLUTION –3 SOLUTION –6 SOLUTION

  13. x log 4 = log 11 4 4 Takelogof each side. 4 x = log 11 x = 11 log 4 x 4 = 11 log 4 x x Solve 4 = 11. log b = x b The solution is about 1.73. Check this in the original equation. ANSWER x 1.73 EXAMPLE 2 Take a logarithm of each side SOLUTION Write original equation. Change-of-base formula Use a calculator.

  14. EXAMPLE 3 Use an exponential model Cars You are driving on a hot day when your car overheats and stops running. It overheats at 280°F and can be driven again at 230°F. If r = 0.0048 and it is 80°F outside,how long (in minutes) do you have to wait until you can continue driving?

  15. Substitute for T, T , T, and r. R ° x x 230 = (280 – 80)e + 80 Ine = log e = x –0.0048t –0.2877 –0.0048t e –0.0048t 0.75 =e –0.0048t 150 = 200e –0.0048t ln 0.75 = ln e 60 t –rt T = ( T – T )e + T R ° R EXAMPLE 3 Use an exponential model SOLUTION Newton's law of cooling Subtract 80 from each side. Divide each side by 200. Take natural log of each side. Divide each side by –0.0048.

  16. ANSWER You have to wait about 60 minutes until you can continue driving. EXAMPLE 3 Use an exponential model

  17. 9x 5. 7 = 15 x 4. 2 = 5 –0.3x 6. 4e –7 = 13 for Examples 2 and 3 GUIDED PRACTICE Solve the equation. SOLUTION about 2.32 about 0.155 SOLUTION SOLUTION about–5.365

  18. log (4x – 7) = log (x + 5). 5 5 Solve log (4x – 7) = log (x + 5). 5 5 ANSWER The solution is 4. EXAMPLE 4 Solve a logarithmic equation SOLUTION Write original equation. 4x – 7 = x + 5 Property of equality for logarithmic equations 3x – 7 = 5 Subtract xfrom each side. 3x = 12 Add 7 to each side. x = 4 Divide each side by 3.

  19. (4x – 7) = (x – 5) ? (4 4 – 7) = (4 + 5) 9 = 9 log log log log log log 5 5 5 5 5 5 EXAMPLE 4 Solve a logarithmic equation Check: Check the solution by substituting it into the original equation. Write original equation. Substitute 4 for x. Solution checks.

  20. Solve (5x – 1)= 3 (5x – 1)= 3 (5x – 1)= 3 4log4(5x – 1) = 4 log log log 4 4 b ANSWER x The solution is 13. b = x EXAMPLE 5 Exponentiate each side of an equation SOLUTION Write original equation. Exponentiate each side using base 4. 5x – 1 = 64 5x = 65 Add 1 to each side. x = 13 Divide each side by 5.

  21. Check: (5x – 1) = (5 13 – 1) = 64 log log log log 4 4 4 4 3 Because 4 = 64, 64= 3. EXAMPLE 5 Exponentiate each side of an equation

  22. log [2x(x – 5)] 2 10 = 10 EXAMPLE 6 Standardized Test Practice SOLUTION log 2x +log(x – 5) = 2 Write original equation. log [2x(x – 5)] = 2 Product property of logarithms Exponentiate each side using base 10. 2x(x – 5) = 100 Distributive property

  23. 2 x – 5x – 50 = 0 log b 2 2x – 10x – 100 = 0 2 2x – 10x = 100 x b = x EXAMPLE 6 Standardized Test Practice Write in standard form. Divide each side by 2. (x – 10)(x + 5) = 0 Factor. x = 10 orx = – 5 Zero product property Check: Check the apparent solutions 10 and –5 using algebra or a graph. Algebra: Substitute 10 and –5 for xin the original equation.

  24. log (2 10) + log (10 – 5) = 2 2 = 2 EXAMPLE 6 Standardized Test Practice log 2x + log (x – 5) = 2 log 2x+ log (x– 5) = 2 log [2(–5)] + log (–5– 5) = 2 log (–10) + log (–10) = 2 log 20 + log 5 = 2 Because log (–10) is not defined, –5 is not a solution. log 100 = 2 So, 10 is a solution.

  25. ANSWER The correct answer is C. EXAMPLE 6 Standardized Test Practice Graph: Graph y = log 2x + log (x – 5) andy = 2 in the same coordinate plane. The graphs intersect only once, when x = 10. So, 10 is the only solution.

  26. 10.log (x + 12) + log x =3 4 4 8.log(x – 6) = 5 2 for Examples 4, 5 and 6 GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 7. ln (7x – 4) = ln (2x + 11) 9. log 5x + log (x – 1) = 2 SOLUTION 5 SOLUTION 3 4 SOLUTION SOLUTION 38

  27. The apparent magnitude of a star is a measure of the brightness of the star as it appears to observers on Earth. The apparent magnitude Mof the dimmest star that can be seen with a telescope is given by the function Astronomy EXAMPLE 7 Use a logarithmic model M = 5 logD + 2 where Dis the diameter (in millimeters) of the telescope's objective lens. If a telescope can reveal stars with a magnitude of 12, what is the diameter of its objective lens?

  28. 2 Log D 10 = 10 ANSWER The diameter is 100 millimeters. EXAMPLE 7 Use a logarithmic model SOLUTION M = 5 log D + 2 Write original equation. 12 = 5 log D + 2 Substitute 12 for M. 10 = 5 log D Subtract 2 from each side. 2 = log D Divide each side by 5. Exponentiate each side using base 10. 100 = D Simplify.

  29. for Example 7 GUIDED PRACTICE 11.WHAT IF?Use the information from Example 7 to find the diameter of the objective lens of a telescope that can reveal stars with a magnitude of 7. SOLUTION The diameter is 10 millimeters.

  30. 7.6 Assignment Hint #3: Quadratic Formula

  31. 6 16 ANSWER ANSWER 5 5 about 0.77 ANSWER ANSWER about 13min ANSWER ANSWER 9 WARMUP Daily Homework Quiz For use after Lesson 7.6 Solve. 1.25x= 125–x + 2 2.8x = 5 3.log7(5x– 8) =log7 (2x + 19) 4.log3 (5x+ 1) = 4 5.log5 5x + log5 (x– 4) = 2 Bonus.Boiling water has a temperature of 212° F. Water has a cooling rate of r = 0.042. Use the formula T = (T0 –TR)e –rt + TRto find the number of minutes tit will take for the water to cool to a temperature of 80°F if the room temperature is 72°F.

  32. Chapter 7 Test Review Assignment p.541: 17-20 all, 24-34 all (16 Q's) Chapter Test p.543: 10-12 all, 16-24 all, 28 (A = Pert) (13 Q's) Do as much as you can without a calculator. Use your calculator to check (if not in the back)

  33. Answers to the Review (p.541) 16 Q's 28. ln 3 + ln y – 5 ln x 29. log7 384 30. ln(12/x2) 31. ln36 32. 2.153 33.7 34. 3.592 • 5 • 0 • -3 • -1/3 • 6.86mm, 5.39mm • log83 + log8 x + log8 y • ln10 + 3ln x + ln y • log8 – 4 log y • ln 3 + ln y – 5 ln x

  34. Answers to the Test (p.543) 13 Q's • 1.732 • 0.874 • 104 • 2 28. $3307.82 • 2 • -5 • 0 16. ln(49/64) 17. log496 • log(5x/9) • 2.431 • 1.750

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